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THE LONDON SCIENCE CLASS-BOOKS 

ELEMENTARY SERIES 

EDITED BY 

PROF. G. C. FOSTER, F.R.S. and SIR PHILIP MAGNUS 



HYDROSTATICS AND PNEUMATICS 



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HYDROSTATICS AND PNEUMATICS 



BY 

SIR PHILIP MAGNUS 

OF THE CITY AND GUILDS OF LONDON INSTITUTE ; 

AUTHOR OF 'lessons IN ELEMENTARY MECHANICS/ &C 

JOINT EDITOR OF THIS SERIES 



EIGHTH EDITION 



LONDON 
LONGMANS, GREEN, AND CO. 

AND NEW YORK: 1 5 EAST 1 6* STREET 
189I 

All rights reserved 

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/{c»4lo I 



HARVARD 

UNIVERSITY 

LIBRARY 

' ^92 



^ 



Richard Clay & Sons, Limited, 
London & Bungay. 



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EDITORS' PREFACE. 



Notwithstanding the large number of scientific 
works which have been published within the last few 
years, it is very generally acknowledged by those who 
are practically engaged in Education, whether as 
Teachers or as Examiners, that there is still a want of 
Books adapted for school purposes upon several 
important branches of Science. The present Series 
will aim at supplying this deficiency. The works 
comprised in the Series will all be composed with 
* special reference to their use in school-teaching ; but, 
at the same time, particular attention will be given 
to making the information contained in them trust- 
worthy and accurate, and to presenting it in such a 
way that it may serve as a basis for more advanced 
study. 

In conformity with the special object of the Series, 
the attempt will be made in all cases to bring out the 
educational value which properly belongs to the study 
of any branch of Science, by not merely treating of its 



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VI Editors' Preface, 

acquired results, but by explaining as fully as possible 
the nature of the methods of inquiry and reasoning by 
which these results have been obtained. Conse- 
quently, although the treatment of each subject will 
be strictly elementary, the fundaro'^ntal facts will be 
stated and discussed with the fulness needed to place 
their scientific significance in a clear light, and to 
show the relation in which they stand to the general 
conclusions of Science. 

In order to ensure the efficient carrying-out of the 
general scheme indicated above, the Editors have 
endeavoured to obtain the co-operation, as Authors 
of the several treatises, of men who combine special 
knowledge of the subjects on which they write with 
practical experience in Teaching. 

The volumes of the Series will be published, if 
possible, at a uniform price of \s, 6d, It is intended • 
that eventually each of the chief branches of Science 
shall be represented by one or more volumes. 

G. C. F 
P. M. 



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PREFACE. 



This Class-Book is intended for the use of those 
pupils in the upper forms of schools who have 
already acquired some elementary knowledge of the 
principles of Mechanics, and are about to commence 
the study of Hydrostatics and Pneumatics. 

In the treatment of the subject of this volume I 
have endeavoured, as far as possible, to combine the 
Experimental with the Deductive method. When- 
ever a law is stated, some explanation is afforded of 
the several experiments by which that law has been 
established ; and whenever a result is deduced, by 
the aid of mathematical reasoning, from more ele- 
mentary principles, the pupil is shown how this 
result may be experimentally verified 

In the hope that this Uttle work may serve as an 
introduction to more advanced treatises on Hydro- 
statics, I have devoted a few pages to the consider- 
ation of the * Flow of Liquids through Pipes and 
Small Orifices'; and, whilst avoiding the mathe- 



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yiii Preface. 

matical difficulties which the fuller treatment of this 
branch of the subject involves, I have endeavoured 
to bring into prominence some of the leading prin- 
ciples connected with it, which recent investigations 
have aimed at establishing. 

To facilitate the use of this text-book in class- 
instruction, the subject-matter is divided into a 
number, of short sections, in which all the more im- 
portant propositions are illustrated by numerical 
examples. To nearly every section is appended a 
set of exercises, progressively arranged, to be solved 
by the pupil. 

My obligations to the published works of different 
writers are acknowledged in the body of the book. 

P. M 

London^ Savile Club : 
September 1878. 



IMs volume can be obtained with or mthout the answers to the 
Exerciser 



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CONTENTS. 



CHAPTER I. 

PRELIMINARY NOTIOKS. 
'.ECTION PAGR 

I. Nature of Fluid Bodies I 

Essential Diflferences between Solid and Fluid 
Bodies. — Change of Gaseous into Liquid State. — 
Experiments of Dr. Andrews. — Viscosity. — The 
Three States of Matter. 

II. Units of Measurement . . . . . . 8 

Units of Length, Area, and Volume. — C.G.S. 
System of Units. — Units of Mass and Weight, 
— Derived Units. — Geometrical Relations. 

Ill Density, — Specific Gravity . . . ^ , 12 

Definitions and Measures of Density and Spe- 
cific Gravity. — Ordinary Definition of Specific 
Gravity. — Tables of Specific Gravity. —Density 
and Specific Gravity of Compound Substances. — 
Exercises. 

CHAPTER IL 

FLUID PRESSURE ON SURFACES IMMERSED. 

IV. Explanation of Terms. — FascaTs Principle . 20 

Intensity of Fluid Pressure. — Variation of Pres- 
sure with Depth. — Direction of Pressure. — Equal 
Transmissibility of Fluid Pressure. — Mechanical 
Appliances — Communicating Vessels. — Exer- 



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X Contents^ 

SECTION PAGf 

V. Whole Pressure on Surface Immersed . . .30 

Definition and Measure of Whole Pressure. — 
Pressure on Base and Sides of a Vessel containing 
Liquid. — Pascal's Vases, — Exercises. 

VI. Centre of Pressure -37 

Definition. — Centre of Pressure of Rectangular 
Area — Of Triangle.— General Expression. — 
Examples. — Exercises. 



CHAPTER III. 

FLUID PRESSURE ON BODIES IMMERSED. 

VII. Resultant Vertical Presmte. — Principle of Arcki- 

medes 44 

Determination of Resultant Vertical Pressure. — 
Principle of Archimedes. — Experiments. — Real 
and Apparent Weight of Bodies. — Relative Gra- 
vitation of a Body in a Fluid. — Exercises. 

VIII. Floating Bodies, — Metacentte . . . . £i 

Principle of Flotation. — Stable and Unstable 
Equilibrium. — Definition of Metacentre. — 
Examples. — Exercises. 



CHAPTER IV. 

SPECIFIC GRAVITY, AND MODES OF DETERMINING IT. 

IX. Application of Principle of Archimedes to the deter- 
mination of the Specific Gravity of Bodies . 61 

Specific Gravity of a Heavy Solid insoluble in 
Water.— Of a Solid that Floats in Water.— Of 
a Liquid.— Of a Solid soluble in Water.— Of 
Gases. — Exercises. 



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Contents. xl 

SECTION PAGE 

X. Other Methods of Determining Specific Gravity.'^ 

Hydrometers ...... 66 

The Specific Gravity Bottle.— Fahrenheit's Hy- 
drometer.— Nicholson's Hydrometer. — Exer- 
cises. 

CHAPTER V. 

THE MOTION OF LIQUIDS. 

XI Liquids momng by their own Weight 7 1 

Definition of Pressure Height or Head of Liquid. 
— Torricelli*s Theorem. — Relation of Velocity 
of Flow to Sectional Area of Vessel. — Vena 
Contracta. — Liquid Flowing through a Pipe of 
Variable Area. — Relation of Velocity of Flow 
to Pressure. — Flow of Liquid through Small 
Orifice. — Effect of Friction on Pressure at Diffe- 
rent parts of a Pipe. 

Xil. Capillarity 86 

Drop Formation. — Experiments. —Surface Ten- 
sion. — Capillary Phenomena, Experiments.-- 
Principles of Capillarity. — Law of Diameters. 

XIII. Diffusion of Liquids 94 

Graham's Experiments. —Crystalloids and Col- 
loids. — Dialysis. — Osmose. — Structure of Li- 
quids. 

CHAPTER VI. 

THE PRINCIPLES OF PNEUMATICS. 

XIV. General Properties of Gases, — Atmospheric Pressure 102 

Definition of Pneumatics. — Expansibility and 
Compressibility of Gases. — Experiments showing 
that the Air has Weight. — Measure of Atmo- 
spheric Pressure. — Torricelli's Experiment. — 
Barometers. — Barometric Corrections. — Abso- 
lute Pressure per Umt Area. — The Siphon. — 
Exercises. 



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xii Contents. 

SECTION ^^®^' 

XV. BoylisLaw . . . • • * * "4 

Experiments for Pressures greater than Atmo- 
spheric Pressure.— Results of Expenments.— 
For Pressures less than Atmospheric Pressure. 
— Dalton's Law. — Graphic Representation of 
Boyle's Law.— Limits of Boyle's Law; Expen- 
ments of Regnault and Despretz.— Relative 
Densities of the Air at different Heights.— 
Use of Barometer in Determining Heights.— 
Exercises. 

XVI. Diffusion of Gases '^9 

Experiments. — Rate of Diffusion. — Graham's 
I^w.— Kinetic Theory of Gases. 



CHAPTER Vn. 

PNEUMATIC INSTRUMENTS. 

XVn. Diving Bell— Pressure Gauges 



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141 



Description of and Problems on Divmg Bell. 
—Manometers.— Compressed Air Manometer. 
—The Siphon Gauge.— Exercises. 

XVni. Air-Pumps 

Parts of a Pump.— Single-Barrelled Air-Pump. 
—Density of Air after a certain number ot 
Strokes.-Difficulty of Working. -Double- 
Barrelled Air- Pump. -Tate's Air-Pump.- 
Magdeburg Hemispheres. - bprengels Air 
Pump.-Condensing Syringe. —Exercises. 

XIX. Pumps for Liquids 5 

Common or Suction Pump. — Tension of 
Piston-Rod.-The Lifting Pump^The For- 
cing Pump. -With Air Vessel.— Fire-Engme. 
Bramah's Press.— Exercises. 



Miscellaneous Problems 



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HYDROSTATICS. 



CHAPTER I. 

PRELIMINARY NOTIONS. 

I. Nature of Fluid Bodies, 

§ I. Hatter. — What we call matter may exist in 
the solid form as iron, wood, and ice, or in the fluid 
form as water, oil, air, and steam. 

§ 2. Essentjial differences between Solid and 
Fluid Bodies. — If we take any portion of a solid body, 
such as a piece of metal, a sheet of glass, or a block of 
wood, one of the first things we observe is that it 
possesses a definite shape, which cannot be changed 
except by the application of a certain amount of force ; 
we may also observe that it occupies, wherever it may 
be placed, the same amount of space. If, however, 
we take a given portion of a fluid substance, such as 
water or air, we find that it possesses no definite shape, 
and that it moulds itself to the form of the vessel in 
which it is contained. Thus if we pour a certain 
quantity of water from one vessel into another, we 
observe that whilst the volume of the liquid remains 

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2 Hydrostatics, 

the same, its shape changes with that of the portion 
of the vessel which it occupies. 

If, again, we endeavour, as in the act of cutting, 
to separate one part of a solid from another, a certain 
amount of pressure, depending on the nature of the 
material, must be exerted ; but if we pass a smooth 
plane surface, such as the blade of a knife or a sheet 
of glass, in the direction of its plane, through a mass 
of fluid, very little resistance is experienced. 

These experiments show that the particles of a 
fluid are mobile^ i.e., they move freely among one 
another, and cohere so feebly that they can be 
separated from oae another by the application of a 
very slight force. We may, therefore, define a perfect 
fluid as a substance the particles of which move freely 
among one another. 

If we take a straight cylindrical lube, open at one 
end and closed at the other, and fit into it a smooth 
rod of some solid substance, such as iron or wood, 
and press the free end of the rod in the direction of 
its length, the pressure applied is transmitted to the 
closed end of the tube without producing any effect 
on its curved surface ; but if the tube contain a fluid 
instead of a solid, and pressure be applied to it by 
means of a piston fitting into the tube, the pressure 
will be felt not only at the base of the vessel, but like- 
wise at all points in its curved surface. This difference, 
which is characteristic of these two states of matter, 
is also due to the fact that the particles of a solid, 
being more or less rigidly connected, are capable of 
holding together under the influence of a certain 
amount of force, whilst those of a fluid, being to a 

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Liquids and Gases. 3 

great extent independent of one another, tend to move 
away in all directions when acted upon by a force in 
any one direction. For this reason fluid bodies under 
the action of gravity cannot rest on a hard horizontal 
surface, otherwise unsupported, as solids do, but re- 
quire in addition lateral support. Hence the follow- 
ing definition : FMid bodies are those which cannot 
sustain a longitudinal pressure, however small, without 
being supported by lateral pressure also. 

§ 3. Two kinds of Fluids. — There are two great 
classes of fluids, called liquids and gases, which are thus 
distinguished : — 

If a given quantity of water, which is liquid, be 
poured into a vessel, it will occupy a certain portion of 
the vessel and no more ; but if a small portion of 
hydrogen or carbonic acid, which are gases, be intro- 
duced into a vessel, however large, the gas will expand 
so that some of it will be found in every part of the 
vessel. 

If we take a cylindrical vessel fitted with ^<^ »• 
a piston and fill it with water, we shall find HT 
that no amount of pressure we can apply j— 1~ 
will sensibly diminish the volume of the 
water; but if, the vessel being filled with 
air, we press down the piston, the volume 
occupied by the gas will be found to 
diminish as the pressure is increased. 

Liquids, when submitted to very considerable pres- 
sure, have been found to undergo some diminution of 
volume, but so little, that for most practical purposes 
they may be considered as incompressible fluids. 
Hence a. perfect liquid may be defined as a mass 

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4 Hydrostatics. 

which is absolutely incompressible and absolutely 
devoid of resistance to change of shape. Matter, 
however, satisfying this condition does not exist in 
nature. 

The chief difference between the liquid and the 
gaseous state of matter consists in this, that whilst a 
given portion of a liquid has a definite volume, but no 
definite shape, a given portion of a gas has neither defi- 
nite volume nor shape, its volume and shape being 
always the same as that of the vessel containing it 

§ 4. Change from Oaseons into Liquid State. — 
Some substances exist at ordinary temperatures both 
in the liquid and gaseous state. Thus we have steam 
and water, and ether both as a liquid and as a gas or 
vapour. Other gases cannot be reduced to the liquid 
state except under the influence of extreme cold and 
great pressure. The temperature at which the change 
takes place, and the amount of pressure required, 
vary considerably. Until very recently, all efforts had 
failed to reduce certain gases to the liquid condi- 
tion ; and consequently these gases were called 
permanent gases, as distinguished fi-om vapours or 
liquefiable gases. But the experiments of MM. Pictet 
and Cailletet, performed in December 1877, have 
demonstrated as a fact what was previously only an 
inference firom analogy, that every gas is the vapour 
of some liquid, and can be reduced to a liquid under 
the necessary conditions of temperature and pressure. 
A few days only after M. Pictet of Geneva had suc- 
ceeded in liquefying oxygen, M. Cailletet of Chatillon- 
sur-Seine liquefied not only oxygen and carbonic 
oxide, but likewise hydrogen, nitrogen, and air. 

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Gaseous and Liquid States. 5 

Although there is thus no absolute distinction 
between permanent gases and vapoiu^, it is convenient 
to use the word * vapour ' to indicate a gas which 
at ordinary temperatures can be reduced to the 
liquid state. 

Experiments by Dr. Andrews have shown that 
gases in changing into liquids can be made to pass 
through an intermediate condition, in which it is im- 
possible to say to which of these two states of matter 
they more exactiy correspond. By enclosing a vapour 
in a tube, and subjecting it, at a very high tempera- 
ture, to a considerable pressure, the vapour can be 
made to pass by imperceptible degrees, i.e. without 
any apparent optical change, into the liquid state. 
For this purpose the vapour must be compressed to 
that volume which it would occupy in the liquid state, 
the temperature being sufficientiy raised to prevent 
liquefaction from taking place. At a particular tem- 
perature, which is known as the critical point, and is 
diflferent for different gases, the tube is found to be 
occupied by a homogeneous fluid which is neither 
a liquid nor a gas, but which changes into one state 
or the other by slightiy lowering or raising the tem- 
perature, the volume remaining constant We thus 
see that liquids and gases are convertible the one into 
the other, and that matter can pass from one state to 
the other without any perceptible gradations. 

§ 5. Viscosity. — Fluids differ very widely with 
respect to their distinguishing characteristic, viz. the 
mobility of their particles. In a perfect fluid the par- 
ticles are supposed to move among one another with- 
out encountering any frictional resistance such as 

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6 Hydrostatics, 

retards the motion of one solid when moving on the 
surface of another. But no fluid exists which fulfils 
this condition. By agitating a fluid, that is, by causing 
one part of it to move against another, heat is gener- 
ated just in the same way as when two sticks of wood 
are rubbed together. This shows that the particles 
encounter frictional resistance to their motion. 

Gases approach more nearly to the definition of a 
perfect fluid than liquids. The latter are found to 
exhibit every variety of difference with respect to the 
mobility of their 4)articles, some approaching to a 
semi-solid condition, and exhibiting properties inter- 
mediate between the liquidity of water and the rigidity 
of ice. Liquids such as treacle, new honey, and tar, 
in which this frictional resistance appreciably interferes 
with the mobility of the particles, are called viscous. 
This viscosity, or * quasi- solidity,' as it is sometimes 
called, is common to all liquids, and exists, though to 
a small degree, in water. To this property is mainly 
due the resistance which a vessel experiences in its 
passage through the sea ; and it can be shown 
that a body, completely submerged, and moving 
with a uniform velocity through a perfect fluid, would 
experience no resistance whatever to its motion. 

§ 6. The Three States of Matter.— If we compare 
together certain typical examples of solid, liquid, and 
gaseous bodies, such as stone, water, and air, we find 
that they exhibit distinct and characteristic properties, 
by which they may be referred to different classes. 
Keeping these differences in view, we have seen that 
we can define a perfect fluid or a perfect liquid, al- 
though we know very well that matter nowhere exists 

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The Three States of Matter, 7 

which fulfils the conditions of these definitions. But 
when we consider the varieties of solid, liquid, and 
gaseous bodies which experience brings under our 
notice, we discover that matter actually exists in all 
intermediate states between these three typical con- 
ditions. Thus we have glue-like liquids of every kind, 
between a clear limpid liquid and a gelatinous or 
semi-solid mass ; and we have vapours in that critical 
condition through which they can be made to pass 
by imperceptible degrees from the gaseous to the 
liquid state. No strict lines of demarcation can 
therefore be drawn between these various conditions 
of matter ; and the solid, liquid, and gaseous states 
may be regarded as only widely separated forms in 
which, under different conditions, the same substance 
may exist 

§ 7. Hydrodynamics defined. — The science which 
treats of the application of the laws of Dynamics to 
fluid bodies is generally known as Hydrodynamics. 
The axioms, or fundamental principles, of the science 
are Newton's laws of motion, which apply equally to all 
branches of Dynamics. The mobility of the particles 
of fluid bodies gives rise to important diflerences be- 
tween their behaviour and that of solid bodies under 
the action of external forces, which render convenient 
the separate treatment of this subject. 

Under the general head of Hydrodynamics are 
included Hydrostatics and Pneumatics, or the study 
of the laws of motion in their application to liquid 
and gaseous bodies respectively. 



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8 Hydrostatics, 

II. Units of Measurement, 

In the solution of problems in Hydrostatics a 
knowledge of the weights of certain volumes of fluid 
is very generally required. It is desirable, therefore, 
to note at the outset the different standards of measure- 
ment which are commonly employed. 

§ 8. Units of length. — The imit of length gene- 
rally used in England is one foot, which is one-third 
of a yard. The imperial yard is an arbitrary measure- 
ment, not derived from any fixed quantity in nature, 
and is defined * as the distance between two marks on 
a certain metallic bar preserved in the Tower of 
London, when the whole has a temperature of 60° R' 

In the French or metric system the stan^rd is 
the metre, * defined originally as the ten-millionth part 
of the length of the quadrant of the earth's meridian 
from the pole to the equator ; but now defined prac- 
tically by the accurate standard metres laid up in 
various depositories in Europe.' The metre is some- 
what longer than the yard, being equal to i '093623 11 
yard, or to 39*370432 inches. The great convenience 
of this system for ordinary purposes is the employ- 
ment of decimal parts or multiples of the metre, to 
represent smaller or larger units. Thus, in any ex- 
pression, if the units represent metres, the tens re- 
present rt^^f^^-metres, the hundreds hecto-mtixt^, and 
so on. In the same way the first decimal place 
represents deci-mttrts, the second cenfi-mtXxtSy the 
third mi//i-mQtres, and so on. Thus 135724 metres 
represents i hecto-metre, 3 deca-metres, 5 metres, 
7 deci-metres, 2 centi-metres, and 4 milli-metres. 

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U7tits of Length and Volume. 9 

In physical investigations the centimetre is now 
generally accepted as the unit of length. This unit has 
been selected by a Committee appointed by the British 
Association, and is recommended for general adoption 
in a work * published by the Physical Society of Lon- 
don. The system of units, based on the recommen- 
dation of this committee, is known as the Centimetre- 
Gramme-Second system of units, and is generally 
referred to as the C.G.S. system. 

I foot = 30*4797 cm. 

When the number of units is very large, it is 
expressed as the product of two factors, one of which 
is a power of 10. Thus 3240000000 is written 
3*24 X 10®, and 0*00000324 is >vritten 3*24 x io~®. 

§ 9. Units of Area. — In England we use, com- 
monly, the square yard, the square foot, and square 
inch. In the metric system we have the sq. metre, 
the sq. decimetre, sq. centimetre, &c. 
I sq. metre =100 sq. decs. = 10,000 sq. centimetres. 

In the C.G.S. system of units, the unit of area is 
the square of the unit or length, i.e. i sq. centimetre. 
I sq. foot =: 929*01 sq. cm. 

§ 10. TTnits of Volume. — The advantages of the 
metric system over that ordinarily used in this country 
are most apparent when we have to compare units of 
volume with units of length. In our system of mea- 
sures no simple relation exists between these two units. 
The gallon, which is the common unit of volume, 
cannot be represented by any exact number of cubic 
feet or inches. In the French metric system the unit 

• * Illustrations of the C.G.S. System of Units,' by J. D. 

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lO Hydrostatics. 

of volume is the litre^ and a litre is equal to a cubic 
decimetre. Thus : — 

I cubic metre = looo litres. 
I gallon = 4*54346 litres = 277*274 cubic inches. 

In the C.G.S. system the unit of volume is the 
cube of the unit of length, i.e. the cubic centimetre. 

I cubic foot = 28316 cubic centimetres. 

§ II. ITnits of Mass and Weight.— The British 
unit of mass is the quantity of matter which weighs 
one pound. It is defined by standard only. 

The French standard is the kilogram, defined 
originally as the quantity of matter in a litre of water 
at 4° C, but now practically determined as the mass 
of a particular piece of platinum preserved in the 
Ministfere de ITntdrieure at Paris, and by standards 
which have been compared with this. 

In the C.G.S. system, the unit of mass is one gram, 
and is equal to the mass of a unit-volume, i.e. a cubic 
centimetre of water, at 4° C. 

I grain = '0647990 grams. 

As the weights of bodies are proportional to their 
masses at places equally distant from the earth's 
centre, the foregoing units of mass may be taken as 
equivalent units of weight. Thus the weight of 3 
cubic centimetres of water is 3 grams, and the volume 
occupied by 3,000 grams of water is 3,000 cubic 
centimetres. 

§ 12. Unit of Time ; derived Units.— The unit of 
time is one second. Units of velocity, acceleration, 
momentum, force, energy, heat, &a, which are based 

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Preliminary Notions. 1 1 

on the fundamental units of length, mass, and time, 
are called derived units. Thus in the C.G.S. system 
the unit of force is that force which, acting upon a 
gram for one second, generates a velocity of one 
centimetre per second. 

§ 13. Geometrical Eelations. — ^As problems fre- 
quently occur which presuppose a knowledge of the 
measurements of the areas and volumes of certain 
figures, the following geometrical relations should be 
remembered : — 

(i) The ratio of the circumference of a circle to 
its diameter = 3-14159 = fjg = \^ nearly, 
and is represented by the Greek letter x. 

(2) The circumference of a circle = 2 tt r, where 

r is the radius of circle. 

(3) The area of a circle = ir r*. 

(4) The area of the surface of a sphere = 4 ir r*. 

(5) The volume or contents of a sphere = | tt ^^. 

(6) The area of the curved surface of a cylinder 

equals the product of the height into the 
circumference of the base = 2 tt rh, 

(7) The volume of a cylinder equals the product 

of the height into the area of the base, 
= TT r'^,h, 

(8) The area of the curved surface of a cone equals 

the product of the slant side into half the 
circumference of the base, = tt rs/h^-^r^y 
where h is the height of the cone. 

(9) The volume of a cone equals one-third of the 

volume of a cylinder on the same base, and 
of the same height, = 3^ t r'^,h. 

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12 Hydrostatics. 

III. Density, Specific Gravity, 

§ 14. Density. — If we take two vessels of equal 
capacity and fill the one loosely with some substance, 
such as sand, and compress into the other a much 
larger quantity of the same substancCy we should say 
that the density of the matter in the one vessel was 
less than that in the other. If, then, we understand 
by mass quantity of matter, we see that the densities 
of two bodies of equal volume and of the same material 
are proportional to their masses. 

§ 15. Measure of Density. — The measure of density 
is the mass of a unit- volume. If we adopt the cubic 
centimetre as the imit- volume, and the gram, or quan- 
tity of matter in a cubic centimetre of water at 4° C., 
as the unit of mass, then the density of a body is 
measured by the number of grams in a cubic centi- 
metre of its substance ; and if //represent the density 
of a body whose volume is V and mass M^ d is the 

M 
mass of a unit volume, and Jf = Vd\ or // = — . 

§ 16. Specific Gravity. — If we take two vessels 
of equal capacity, and fill the one with mercury and 
the other with water, we shall find that the one con- 
taining the mercury is heavier than the one filled with 
water. This difference in the properties of the two 
substances is known as a difference in their specific 
gravities. When we say that lead is heavier than 
wood, we mean that bulk for bulk the one substance 
is heavier than the other — that a cubic foot of lead 
weighs more than a cubic foot of wood. 

§ 17. Measure of Specific Gravity.— The sp. gr. 

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Density and Specific Gravity, 13 

of a substance is measured by the weight of a unit- 
vohime of that substance. If d be the mass of a unit- 
volume, and s its weight, then s=:gd^ where ^ is the 
acceleration due to gravity ; and if W^ be the weight 
of a body whose volume is V and sp. ^. s, then 

W=Vs,oxs=: ~ 

Since, also, s = gd, we have JV^gd, V, 

If we adopt the weight of a gram as the unit of 
weight, the specific gravity of a body is expressed by the 
weight in grams of a cubic centimetre of its substance. 

Ordinary definition of Specific Gravity. — The spe- 
cific gravity of a substance is very often said to be 
measured by the ratio of the weight of a given volume 
of that substance to the weight of an equal volume 
of some standard substance ; and in considering solid 
and liquid bodies, water at 4° C. is taken as the stan- 
dard ; whilst in the case of gases, air at 0° C. and 76 cm. 
barometric pressure is employed. But it will be seen 
that by defining specific gravity as the weight of a unit- 
volume, we avoid the explicit reference to a ratio, whilst 
the number expressing the ratio, when water is the 
standard substance, is the same as the number of 
grams representing the specific gravity. Thus, if 
W be the weight of a given volume V of any sub- 
stance, and W the weight of the same volume V of 
water, then according to the ordinary definition 
W -^ W '=' specific gravity. But if s be the weight 
of a unit- volume of the substance, and a/ the weight of 
a unit- volume of water, W^=^ Vs and W^ = Viv. 

Vs s 
Hence, specific gravity = -- - = — , and if wc take 

Vw w 

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14 Hydrostatics. 

the weight of a unit- volume of water to be the unit 
weight, as we have supposed, then the specific gravity 
of a substance = j, the weight of a unit-volume of 
that substance. 

Since, also, the weight of a unit of mass is equal to 
g units of force, and is represented in gravitation units 
by one gram, we see that the numbers representing 
the densities of bodies represent also their specific 
gravities, and that the specific gravity of water is unity. 

When the specific gravity is considered as a ratio, it 
is sometimes called the relative specific gravity y to dis- 
tinguish it from the absolute specific gravity ^ or weight 
of the mass of a unit- volume. 

Where, as in the English system of weights and 
measures, the weight of a unit- volume of the standard 
substance is not adopted as the unit of weight, the 
specific gravity of any substance, considered as a 
ratio, = J" -f- a/, where s is the weight of a unit- volume 
of the substance, and w is the weight of a unit- volume 
of the standard. If, therefore, S be the relative 
specific gravity of the substance, we have ^5= Vs-^w 
or VSw =zVs = lV. 

Thus, if we wish to find the weight of 4 cubic 
inches of zinc, we must first know the weight of a 
cubic foot of water (a/), and then the specific gravity of 
zinc (represented by a ratio) being 7, we have JF= 
7 X ttVb- ^ ^y ^^d taking w to equal 1,000 oz. roughly, 
we have H^= 7X4xiooo ^ ^.^^ 1^^^ ^^^^ 
16x1728 

The weight of a cubic foot of water is more nearly 
equal to 62*3 lbs. 

In estimating the weights of gases, it is useful to 

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Tables of Specific Gravity. 



^S 



remember that one cub. cm. of dry air at o° C. and 
76 cm. pressure (at Paris) weighs 0*001293 grams. 
Under the same circumstances, 13 cubic feet of air 
weigh very nearly i lb. avoirdupois. 



§ 18. SPECIFIC GRAVITY OF SOME IMPORTANT 
SUBSTANCES. 

TABLE I. 
Solids. 



Name of Substance 



Specific 
Gravity 



Name of Substance 



Specific 
Gravity 



Platinum, cast 
Gold, cast . 
Lead, cast . 
Silver . . . 
Bismuth . . 
Copper, hammer^ 

„ wire 
Brass . . 
Nickel . 
Steel . , 
Iron, wrought 
Iron, cast 
Tin . . 
Zinc . . 
Antimony 
Iodine 
Diamond . 
Flint-glass 
Aluminium 
Bottle-glass 
Plate-glass 
Marble . 
Emerald . 
Rock-crystal 
Porcelain 



20-86 

1925 

11-35 

10-47 

9-82 

8-88 

878 

8*39 
8-28 
7-82 
779 

7*21 

7-29 

7 'CO 

671 
4*95 
3-52 
378 to 3-2 
2-57 
2-6o 

2-37 
2-84 
277 
2-66 
2*49 to 2*14 



Sulphur, native 
Ivory .... 
Graphite . . . 
Phosphorus . . 
Magnesium . . 
Amber . . . 
Wax, white . . 
Sodium , . . 
Potassium . . 
Ebony, American 
Oak, English . 
Mahogany, Spanish 
Box, French 
Beech . . 
Ash . . 
Maple 
Cherry-tree 
Walnut . 
Pitch pine 
Elm . . 
Cedar 
Willow . 
Larch . . 
Poplar 
Cork . . 



203 
1*92 
1*8 to 2*4 
177 

174 
I -08 
097 
0-97 
0-86 
1*33 
o'97 to I '17 
I '06 
1*03 
0-85 
0-84 
075 
071 
0-68 
0-66 
060 
059 
0-58 
0-54 
038 
0*24 



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i6 



Hydrostatics. 



TABLE II. 
Liquids, at o® C. 



Name of Substance 


Specific 
Gravity 


Name of Substance 


Specific 
Gravity 


Mercury .... 
Sulphuric acid . . 
Nitric acid . . . 
Aqua regia . . . 
Hydrochloric acid . 
Blood, human . . 
Ale, average . . 

Milk 

Sea-water . . . 
Vinegar .... 
Tar 


1-848 
1*500 

1*234 
I -218 

1*045 
1*035 
1-030 
I '028 
1-026 
I 015 


Water, distilled, at 

Linseed oil . . . 
Proof spirit . . . 
OHve oil ... . 
Ether, hydrochloric 
Turpentine, oil of . 
Brandy .... 
Alcohol, absolute . 
Ether, sulphuric . 


I 000 
0-940 
0-930 
0-915 
0-874 
0870 

0-837 
0796 
0720 



TABLE III. 
Gases at 0° C. and 76 cm. Pressure. 



Name of Substance 


Specific 
Gravity 


Name of Substance 


Specific 
Gravity 


Oxygen .... 
Atmospheric air 
Nitrogen .... 
Hydrogen . . . 
Chlorine .... 


0001432 
0001293 
0001267 
0000894 
OX)03209 


Hydrochloric acid 

gas 

Nitrous oxide . . 
Carbonic acid . . 


0-00164 
0-00197 
0-00198 



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Density and Specific Gravity of Compounds, 17 

§ 19. To find the density of a combination of 
two or more substances whose volnmes and densities 
are ^ven. 

Let z/j, z/2, v^ ... be the volumes of the substances, 
d\^ ^2> ^3«- their respective densities. 

Then if Fbe the volume of the combination, and 
if no contraction take place, 

V=Vi +V2+ Vq 4- ... 

and if Z) be the density of the whole, V£> equals the 
mass of the whole, and therefore, 

VD = Vi di +^^2 ^2 + ^'3 ^3 + ••• 

or Z) = i^lAjt" ^2 ^2 + z^3 ^8 + '" 
Vi + >2 + v^ +... 

If, however, as very frequently happens, contraction 
takes place, and if the volume of the whole is some 
proper fraction (= r) of the sum of the volumes of 
the parts, 

v,d, ^ v^d^^v^d^ + ... 
^^^■^-r(z/i + z;, + z/3...) 

A similar proposition holds good if, for density, 
we substitute specific gravity. 

§ 20. To find the specific gravity of a combina- 
tion of substances, the weights and specific gravities 
of which are given. 

Let ze/i, 0/3, a/3 ... be the weights of the components, 
and ^1, J2, s^ ... their respective specific gravities ; then, 
if W be the weight of the whole and S its specific 
gravity, we have IV = w^ + 0/3 + 0/3 4- ... 

G 

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iS Hydrostatics. 

and since W = VS^ where V is the volume of the 
whole 

i^ — . ^ -4- -^ 4- J£a 4- 
O Jj ^2 "^3 

supposing no contraction to take place, and there- 
fore 

«$"! '$'2 "^3 

But if the volume of the whole be less than the siim 
of the volumes of the parts in the ratio of r : i, then 

2— ^\ 4- «^2 + ^3 + ... 

^\S^ ^2 ^3 ^ 

The density of a compound can be found in the 
same way, the masses and densities of the components 
being given. 

§ 21. Examples.— (I.) i,ooo cc. of a gas whose density 
is 12 are mixed witb 2,cxDO cc. of a gas whose density is i6, 
and the volume of the mixture is diminished by one-third. 
Find the density of the mixture. 

T^ 1000x12 + 2000x16 

U B _ = 22 

f(l000 + 2000) 

(2.) Three kilograms of a substance, sp. gr. = 4, are melted 
with 5 kils. of a substance, sp. gr. = 6, and the volume of the 
mixture is o*i less than the sum of the volumes of its com- 
ponents. Find sp. gr. of mixture. 

^ = 3 + 5 _ 8x240 _ -35 
&(* + g) "" 9x38 '^^ 

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specific Gravity of Compounds, 19 



Exercises. I, 

1. Find the weight of a bar of copper, the section of which 
is 64 square centimetres, and length one metre. ^ 

2. Find the size of an iron shot that weighs 60 lbs. ^/ v, c^c - 

3. Find the weight of a column of mercury 30 inches high, 
having a uniform section of one square inch. 1 1^ r* 

4. A cubic inch of a substance weighs 2 oz. ; find its specific 
gravity. J. ^ 

5. If silver and copper are mixed in the proportion of 2 : 7 
by weight, find the specific gravity of the compound f y Q 

6. What mass of copper must be mixed with 200 grams of 
gold to make the specific gravity of the compound 0*9 of that 
of gold? >.:^,^^W.-^ 2,/. /)3*w- 

7. Equal weights of two fluids, of which the specific gravities 
are s and 2j, are mixed together, and the mixture occupies three- 
fourths of the sum of the volumes of its components. Find the 
specific gravity of the mixture. ^)^ 

8. Three litres of a gas, the density of which is unity, are 
mixed with one litre of a gas density 14, and the mixture occu- 
pies half the volume of the original gases. . Find its density. ?i-^ 

9. A bar of cast-iron, sp. gr. = 7*2, is found to weigh 17-5 
kils. The length of the bar is 5 decimetres, and sectional area 
50 sq. centimetres. Is there a flaw in the casting ? If so, what 
Is its size? l^Qj^ ^^^ 



C3 

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20 Hydrostatics. 



CHAPTER 11. 

FLUID-PRESSURE ON SURFACES IMMERSED. 

IV. Explanation of Terms. PascoTs Principle, 

§ 22. Fluid Pressure. — The base and sides of a 
vessel containing a liquid in equilibrium are subjected 
to a certain pressure, which is caused by the weight 
of the liquid, and by the resistance which the surface 
of the vessel offers to the free motion of the liquid. 

§ 23. Intensity of Pressnre.—By intensity of 
pressure at a point is meant the pressure on the unit 
of area containing that point 

If a small surface a^ in contact with a fluid, is 

maintained in equilibrium by a force jP, acting perpen- 

p 
dicularly to the surface, then - measures the average 

a 

pressure-intensity at any point of the area ; and if 

P 

-= /, then/ is the pressure on a unit of area. 
a 

§ 24. Variation of Pressure with DeptL— Sup- 
pose the small horizontal area a to be at a depth i 
below the surface of the liquid, then the area a will be 
pressed vertically downwards by a force equal to the 
weight of the column of liquid above it ; and if x equal 
the weight of a unit volume of liquid, this force is equal 

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Variation of Pressure with Depth, 21 

io z as. If, then, / be the pressure at any point of 
this area, p .a^= zas, orp = z s. 

If a is not horizontal, the pressure at some parts 
of the area is greater than at others. But if we sup- 
pose this area to become smaller and smaller, and 
ultimately to diminish without limit, the difference of 
pressure at different parts of the area will diminish 
likewise ; and if / represent the pressure intensity, 
when the area has become so small that it may be 
regarded as a point, then/. a^=z.a. s, where a is the 
infinitely small area about this point, and z is its 
depth. Hence/ = z,s, as before, showing that /;/ t/ie 
same liquid t/ie pressure varies with the depth. 

This proposition may be experimentally verified by 
taking a small cylindrical vessel a (fig. 2), open at 
both ends, and having a movable base o, to which a 
thread c is attached. If we connect the base with one 
end of a scale-beam and press y\c, a. 

it against the vessel by at- 
taching a weight W to the 
other end of the balance, we 
shall find, on pouring water 
into the vessel, that the disc 
falls off when the water has 
risen to a certain height, and 
that this height is always pro- 
portional to the excess of the 
weight W above the weight ' 

of the disc. This shows that 

the ratio of the pressure on the disc to its depth below 
the surface of the liquid is constant, />. that the pres- 
sure varies with the depth. 

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22 Hydrostatics, 

By fixing the vessel a into another vessel contain- 
ing water, it may be shown that the liquid in the outer 
vessel exerts an upward pressure on the under side 
of the disc equal to the downward pressure of the 
liquid contained in the inner vessel. For, if we attach 
the disc o to one end of the scale-beam, and to 
the other end a weight sufficient to counterbalance it, 
we shall find, on pouring water into a, that the disc 
does not fall off until the level of the water in the two 
vessels is the same, 

§ 25. Direction of Pressure on a Surface in Con- 
tact. — If a thin plate be immersed in a fluid and 
be held in any position, the direction ot the pressure 
is perpendicular to the surface of the plate, if the 
fluid is at rest. For if it acted in any other direction 
it could be resolved into two components, one per- 
pendicular to the surface and the other along it; 
and the latter component would, in the absence of 
friction between the surface and the fluid, produce 
motion, which is contrary to supposition. The direction 
of the pressure must, therefore, be perpendicular to 
the surface. 

§ 26. Equal Transmissibility of Fluid Pressure. 
If we take a vessel full of water (fig. 3), having 
various apertures of the same size, fitted with water- 
tight pistons which are kept in equilibrium, and 
if one of these a be pressed downwards with a force 
p, an additional pressure equal to p will be required 
at each of the other pistons to preserve equilibrium. 
Thus if a force of i lb. be applied at a, it will be 
necessary to apply an increased pressure of i lb. to 
each of the other pistons to prevent motion ; and as 

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Equal Transmissibility of Fluid-pressure, ^3 
it matters not in what part of the vessel these pistons 



Fig. 




are fitted, we see that any 

increase of pressure applied 

at one part is transmitted 

equally throughout the fluid. 

If the pistons, as in fig. 4, 

are of different areas, that 

of B being twice that of a 

and the area of c being 

nine times as great, it will 

require an additional force 

of 2 lbs. at B, and of 9 lbs. 

at c, to preserve equilibrium, when a force of i lb. 

IS applied at a. 

These experiments serve to 
illustrate the following fundamental 
law of fluid-pressure, which is 
known as Pascal's ^ principle : 

When pressure is communicated 
to any part of a fluids it is trans- 
mitted equally in all directions 
through the fluid. 

§ 27. Hechanical Application. 
principle, which is a ne- 
necessary consequence of 
the mobility of the par- 
ticles of a fluid, serves to 
explain the action of a 
very useful mechanical 
contrivance for multiply- 
ing power. 

* Pascal was bom at Auvergne, 1623 ; dkd 1662. 

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24 Hydrostatics. 

It consists of two communicating vessels contain- 
ing water, one of which is much larger than the 
other. The vessels are fitted with pistons P and 
/, the areas of which we will suppose to be A and 
a. If now weights ?f^ and zc' be placed on those two 
pistons respectively, so as to counterbalance one 
another, it will be found that W \w\\ A : a, which 
is in accordance with Pascal's principle. 

We see, also, that if the piston / be pressed down 
through the space S, the water contained in the 
smaller vessel will pass into the larger, and force up 
the piston F through some space s, such that— 

ax SssAxs, 
since the volume of water that is removed from one 
vessel is the same as that which enters the other 
vessel. Hence 

S^A_W 
saw 
or, w, S=i W, s, 

/>., the work done by 7V = work done by W. 

§ 28. Hydrostatic Paradox. — A consequence of 
Pascal's principle is that a quantity of water, however 
small, can be made to support a weight, however large, 
and this seeming paradox can be exhibited in the 
following manner. 

Let A B (fig. 6) be a long narrow pipe communi- 
cating with a vessel c d, into which a piston c e is 
fitted. If now water be poured through the pipe, it 
will be found to rise to the same level, c e f, in 
both parts of the vessel ; but if a weight Wht placed 
on c E, and additional water be poured into the pipe, 

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Flow dependent on difference of Surf ace4eveL 25 




a position of equilibrium will result such as is shown 

in the figure. In this case, the pressure exerted by 

the water in the pipe above 

the plane c e f supports the 

weight W. Since this pressure 

is communicated to every 

element of area of c e equal 

to the sectional area of the 

pipe, the whole pressure on 

c E is as many times greater 

than the weight of the water 

in A F as the area of c e is 

greater than the area of the 

pipe. 

Let a equal the sectional area of the pipe ; then if 
P is the pressure at f produced by the weight of water 
above it, /^ = dJ x a f x a/, where w is tl\e weight of 
unit-volume of water, and if A is area of piston c E, 
the pressure communicated to c e is 

— XaXAF XW= AXAFXW 

a 

/. W =iA X A F X a/. 

This shows that the weight IV can be made as great 
as we please, by taking A f or c e sufficiently great, 
and that it is independent of «, the section of the tube. 
Hence, with a tube sufficiently narrow, a quantity of 
liquid, however small, can, in principle, support a 
weight, however large. 

§ 29. Communicating Vessels containing Liquid. 
If two vessels a and b (fig. 7) containing the same 
liquid be connected by a pipe, and if the surface level 

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26 Hydrostatics, 

of the liquid in a be higher than that in B, the liquid 
will be found to flow from a to b. 

This is easily explained by considering the pres- 
sure on either side of any sec- 
tional area, / ^, of the pipe. 
For it is evident that the pres- 
sure on the side towards A is 
greater than on the side to- 
wards B, since the pressure 
varies with the depth below 
the free surface, and consequently the flow takes place 
from A to B. 

It should be observed that although in the case 
now considered the liquid flows from a to b, the 
pressure-intensity along the base of a is less than 
that along the base of b , and also that the pressure 
at a^ where the fluid escapes from the one vessel, is less 
than that at ^, where it enters the other vessel. 

If we connect by a tube c d two parts of the 
same vessel, although the pressure at d is greater than 
that at c, no flow takes place, for it will be seen that 
the pressure on either side of any element of area in 
the pipe c d is the same. 

We see, therefore, that a liquid always flows from 
places of higher to places of lower surface level, and 
that if a liquid is in equilibrium its surface level must 
be uniform. Hence it follows that if a liquid be con- 
tained in a vessel, its surface must be horizontal ; or, 
more generally, every point of the surface of a liquid 
in equilibrium must be at the same distance from the 
earth's centre. In the case of large inland seas, the 
surface is somewhat curved. 

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Communicating Vessels, t.'j 

We also see that the pressure at all points in the 
same horizontal plane within a liquid at rest must be 
the same, since these points are all equally distant 
from the free surface, and the pressure varies with the 
depth (§ 24). 

It also follows that if any number of vessels con- 
taining liquid communicate with one another, the 
liquid will stand at the same level in all. For if the 
surface level were higher in any one vessel, a flow of 
liquid would take place from that vessel into the 
others, till the uniformity of surface-level had been 
established.^ 

This fact may be experimentally verified by pour- 
ing water into one of the parts fig. 8. 
of a vessel similar to that ^^^ M, £/ £f 
shown in the figure, when it ^^ ^H S M 
will be found that the free sur- |^^^^^^^^gi 
faces of the liquid in all parts ^ 
of the vessel lie in the same horizontal plane. The 
tendency of liquids to find their own level, />. 
to flow from places of higher to places of lower 
surface-level, is of great practical importance. It 
is utilised in the water-supply of towns. A reser- 
voir of water is kept at a considerable elevation, and 
from it pipes proceed in all directions conveying water 
to any heights below the level in which it stands in 
the main reservoir. The water-level is an instru- 
ment which acts on the same principle. It consists 
of a tube bent at right angles, and furnished at its 

^ It should be observed that difference of surface-level cor- 
responds with difference of temperature in heat, and with 
difference of potential in electricity. 

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28 Hydrostatics. 

extremities with two glass vessels, and the whole is 
partly filled with water. Since the liquid has the 
same surface-level in both branches, any points which 
the observer's eye detects to be on the same level 
with the surface of the water in both branches, will 
be in the same horizontal plane. 

When a liquid falls from a higher to a lower level, 
there is a change of potential into kinetic energy, 
and a corresponding amount of work can be effected 
This fact is economically employed in water mills. 



§ 30. Examples. — (i.) Find the whole pressure exerted on 
a horizontal area of 9 sq. cms. which is sunk 125 cms. below 
the surface of water. The pressure equals the weight of water 
supported, 

= 125x9=1125 grams. 

(2.) Find the pressure- intensity due to a column of 25 cms. 
of mercury upon which rests a column of 50 cms. of water. 

The pressure-intensity is the pressure per unit area, i.e, per 
sq. cm. : 

= 25 X 13 '6 + 50 = 390 grams = 390 g units of force, 

the specific gravity of mercury being 13*6. 

(3.) Find the vertical force necessary to support the hori- 
zontal base of a vessel containing mercury, if the area of the 
base is one square decim., and its depth below the surface of the 
mercury 5 centimetres, neglecting the weight of the base. 

°=^ d! j« 5 X 100 X 13*6 grams. 
.•. P^ 6800 grams. 

(4.) What must be the height of a column of mercury to 
exert a pressure of 1220 grams per sq. centim. ? 



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Fluid-pressure on Surfaces immersed, 29 

If A be the height required, I220s^ x 13*6 grams. 

. , 1220 o_^ 
. , h = — - = 897 cm 

136 

§ 31. Equilibrium of two different Liquids in 
Conmnmicating Tubes. — If two liquids that do not 
mix meet in a bent tube, or in two tubes communicat- 
ing with each other, the heights of their free surfaces 
above their common surface are inversely proportional 
to their specific gravities. 

Let A a' be a plane drawn through the common 
surface of the two liquids in one of the tubes. Let s and 
^ be their respective specific gravities. Let ff and c be 
the free surfaces of the liquids. Then p^^ 
if the liquids are in equilibrium, the 
pressures at a and a' must be the same. 

The pressure-intensity at a is « ^ x jt, 
where ab \% the vertical height of b 
above a. 

The pressure-intensity at a' is « ^ x j' 
^vhere acvst vertical height of c above a'. 

Hence ab y. s ^ ac x s' 

or ab : ac W s' : s. 



Exercises. II. 

1. Two communicating vessels contain fluid, and are fitted 
with pbtons, the diameters of which are 2 inches and 8 inphes 
respectively. If a weight of 3lbs. is placed on the smaller 
piston, what weight must be placed on the larger to preserve 
equilibrium ? ' ' 

2. A narrow vertical pipe is attached to a vessel (fig. 6), 
Which is fitted with a piston the area of which is 2 square deci- 



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30 Hydrostatics, 

metres. If the vessel and pipe contain water, find the height 
of the water in the pipe when a weight of 40 kils. is placed on 
the piston, '^ \ ^ 

3. A cylindrical vessel contains mercury to the height of 2 
inches above the base, and a layer of 8 inches of water resting 
on the mercury. Find the pressure at any point in the base, 
taking sp. gr. of mercury to be 13 '6. 

4. At what depth below the surface of a lake is the pressure 
intensity 5 times as great as at a depth of 10 feet, supposing 
the atmospheric pressure to be equal to the weight of a column 
of water 34 feet high ? 

5. Two liquids that do not mix are contained in a bent 
lube ; the difference of their levels is 3 ins., and the height of 
the densenabove their conmion surface is 5 ins. ; compare their 
specific gravities. 

6. If, in the above, the internal section of the tube is one 
square inch, and the lighter liquid is water ; find the weight of 
water contained in the tube. 



V. Whole Pressure on Surface immersed, 

§ 32. Whole Pressure. — When a vessel contains 
a fluid, or when a body is immersed in a fluid, the 
fluid exerts a normal pressure at each point of thesur- 

FlG. 10. 



face in contact with it. The sum of all these pressures is 
called the whole pressure on the surface immersed. 

It should be observed that these pressures act in 
(Jifferent directions, the pressure at each point being 
perpendicular to the surface at that point The whole 



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WItole pressure on a Surface immersed. 31 

pressure is the sum of all these pressures, and repre- 
sents the total strain to which the vessel containing 
the fluid, or the body immersed, is exposed. 

§ 33. To find the Whole Pressure which a 
Liquid exerts on a Surface immersed. — Let a b c d 
be any area in contact with a fig. 11. 

homogeneous liquid, the free 
surface of which is the horizon- 
tal plane abed. Let ef be 
any element of area so small 
that the pressure-intensity at 
all points may be supposed 
to be constant, and let the 
depth of this area below the 
surface of the liquid be z. Then if s be the weight of 
a unit- volume of the liquid, the pressure exerted one f 
\% a z s^ where a is the area of ^/ 

Now the whole pressure is equal to the sum of the 
normal pressures on all the elements that make up the 
whole area. Therefore the whole pressure is equal to 

ai Zi s + a2 Z2 s -^ a^ z^ s + . . . 
= {^1 Zi -\- a^ Z2 + a^ z^ -\r . . .) s. 

But, by the properties of the centre of gravity, 

a^ Zi + ^2 ^2 + ^3 ^3 + • • • = (^1 + ^2 + ^3 + • • •) ^ 
where z is the depth of the centre of gravity of the 
whole area below the plane surface of the liquid ; 
. '. whole pressure ^= A z s, 

where A is the whole area in contact, z the depth of its 
centre of gravity below the surface of the fluid, and s 
the weight of a unitvolume of the liquid. Or, the 



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32 Hydrostatics. 

pressures Azdg where// is density of fluid and g the 
acceleration due to gravity. Hence, The whole pres- 
sure on any area immersed equals the weight of a column 
of liquid which has that area for base, and the depth 
of its centre of gravity below the surface of the liquid for 
height. 

This proposition holds good whatever may be the 
shape of the vessel containing the liquid ; and what- 
ever may be the position of the area immersed. In 
some cases the results to which it leads seem at first 
sight paradoxical, but they will be shown to agree 
with the general proposition which we have now esta- 
blished. 

§ 34. Examples. — (i.) Compare the pressures on the base 
aiid side of a cube filled with liquid. 

Let the edge of the cube be a, then area of the base is a\ 
and depth of centre of gravity below surface is £Z, 

/. whole pressure on base is a* x a x s^c^s. 

The area of a side is also a^ ; but depth of its centre of gravity is 

~, /, pressure on side istf^x — x j«— j, 
2 22 

,'. whole pressure on base =» twice the whole pressure of one 
of the sides. 

The pressure on the base acts vertically, and the pressure on 
the sides horizontally. 

(2.) An oblong, the edges of which are 6 metres and 8 metres, 
is immersed vertically with its shorter edge horizontal, and 2 
metres below the surface of water ; find the whole pressure on 
either side. 

Area immersed is 6 x 8 = 48 square metres. 

Depth of centre of gravity is 4 + 2 = 6 metres. 

/, whole pressure is 48 x 6 x j, where s is the Weight of 
a cubic metre of water, i,e, 1,000 kilograms. 
/. Pressure - 288,000 kils. 

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Fluid-pressure on Surfaces immersed. 33 



Fig. 




(3.) A right pyramid, the height of which is 8 decs., has a 
square base, each edge of which is 5 decs. Required the whole 
pressure on the base when the pyramid is full of liquid. The 
area of the base is 25 sq. decs., and the 
depth of its centre of gravity below the 
sur&ce of the liquid is 8 decs. Hence 
the whole pressure on the base is 80 
X250OXJS20OX kils., where s is the 
sp. gr. of the liquid. It is to be observed 
that the pressure on the base, in this 
case, is greater than the weight of the 
liquid the vessel contains, being equal 
to the weight of liquid in the prism hav- 
ing the same base, ue, to a column of 

liquid the height of which is E F, and base A B c D. On the 
other hand, the pressure transmitted to the stand on which the 
vessel rests is equal only to the weight of the liquid contained 
in the vessel ; and consequently the pressure on the base of the 
vessel is greater than the pressure communicated to the stand. 
This result which follows directly from the general proposition, 
requires further explanation, and will be considered in the fol- 
io viring paragraph. 

§ 35. To find the whole pressure exerted by a 
liquid on the base and sides of the vessel con- 
taining it. — ^We have three cases to consider, ac- 
cording as the sides of the vessel are vertical or 
slant from the base outwards or inwards. We shall 
suppose the base of the vessel to be horizontal If 
the sides are vertical, as in fig. 13, the pressure on the 
base is evidently equal to the weight of the fluid con- 
tained in the vessel, and the horizontal pressures are 
equal 

Suppose now that the sides are inclined, outwards 
in fig. 14 and inwards in fig. 15. 

The pressure at any point o in the side a b is a 

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.^4 



Hydrostatics, 



force p acting perpendicularly to a b. As this force 
is prevented by- the resistance of the slant side from 






Fia 


15. 


E 


-.^-U 


r 


r 


-^* 
^--j 





causing motion, it produces a reaction equal in mag- 
nitude but opposite in direction. This reaction / 
can be resolved into two components, x and y acting 
at o, X horizontally in both figs., and y vertically up- 
wards in fig. 14 and I vertically downwards in fig. 15. 
It thus appears that in fig. 14 part of the weight 
of the liquid is supported by the sum of the forces 
y acting at all points in the slant side, and the 
remainder of the weight of the liquid presses on the 
base B c. Hence, in this case the pressure on the 
base is /ess than the weight of the contained liquid. 

But in fig. 15 the force y acts downwards, and 
consequently increases the pressure on the base caused 
by the weight of the liquid ; and hence the pressure on 
the base of the vessel is greater than the weight of the 
contained liquid by the sum of the vertical components 
of the reactions caused by the pressure of the liquid 
against all points in the slant sides. 

Again, since the force /, due to the pressure of 
the liquid, and the components, x and y of the reac- 
tion of the surface a b, are in equilibrium, they can 



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Pascal's Vases. 35 

be represented by the sides of the triangle a e b. In 
the same way the forces/', ^',y, acting at q can be 
represented by the sides of the triangle d f c 

Hence/ :^:;c::ab:ae:eb and 

Z' : J'' : ^ : : c D : D F : F c 

But E B is equal to f c, when the base is horizontal 
.-. X = x' \ and in the same way the horizontal 
pressures at all other points can be proved to be 
equal. Hence the horizontal components of the pres- 
sure of the liquid against the slant side are equal 

If the base is not horizontal, then the whole pres- 
sure on the base can be resolved into vertical and 
horizontal components, and the algebraic sum of the 
horizontal pressures on the base and sides of the 
vessel will still be found to equal zero. 

§ 36. Experimental verification. Pascal's Vases. 
— In order to verify the general proposition, which we 
have now proved, viz., that the pressure on the base of 
a vessel containing liquid is independent of the shape 
of the vessel, and varies only with the area of the base 
and the depth of its centre of gravity below the sur- 
face of the liquid, Pascal contrived an experiment very 
similar to the following : 

Take three vessels . p, q, m, of different shapes, 
having the same circular aperture at their base, 
and capable of being screwed into the ring of the 
stand A B. One of the vessels being fastened to the 
ring, a circular disc d hanging from one arm of the 
balance is pressed against it by weights placed in the 
scale-pan hung to the other arm. Water is then 
poured into the vessel, and an index / marks the 



D 2 



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36 Hydrostatics. 

level at which the water stands when the disc falls 
off If the experiment is now tried with the other 

Fig. x6. 




vessels, it is found that the disc falls off when the water 
rises to the same level. This shows that the pressure 
on the base of each vessel is the same when the water 
is at the same height above it 

Exercises. III. 

1. Find the whole pressure on a rectangular surface 6 feet 
by 4 feet, immersed vertically in water with the shorter side 
parallel to, and 2 feet below the surface. 

2. Find the whole pressure on the curved surface of a 
vertical cylinder which is filled with a liquid, sp. gr. = I '5, the 
height of the cylinder being 2 decims., and the radius of the base 
7 cms. 

3. Show that if a sphere or a cube be filled with liquid the 
total strain to which it is subjected is three times the weight of 
the liquid it contains. 

4. A flood-gate is 6 feet wide and 12 feet deep. What is 
the total pressure on the flood-gate when the water is level witir 
the top? 

5. A globe, tha radius of which is 3*5 cms., rests at tho 

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Centre of Pressure. 3 7 

bottom of a vessel 30 cms. in height, foil of water. Find the 
total pressure on the globe. 

6. A conical vessel 2 decimetres high, has a movable base, 
of 25 sq. cms. area, formed by a disc 2 cms. thick, and the 
specific gravity of the material being 3 '2, find the force neces- 
sary to uphold the disc when the vessel is full of water. 

7. A smooth vertical cylinder, 2 feet high and i foot in 
diameter, is filled with water, and closed by a piston weighing 
3 lbs. Find the total pressure on the curved surface. 



VI. Centre of Pressure. 

§ 37. Definition. — When a plane surface is im- 
mersed in a fluid, the pressures at different points of 
the surface are perpendicular to it (§ 25), and consti- 
tute a system of parallel forces of which the whole 
pressure is the resultant. The magnitude of this re- 
sultant pressure we have seen how to find, but the 
point at which it *acts we have not yet determined. 
This point is called the centre of pressure^ and may be 
defined as the point of action of the single force equiva- 
lent to the whole pressure exerted by a fluid on any plane 
surface with which it is in contccct 

If a plane surface is immersed horizontally the 
centre of pressure corresponds with the centre of gravity, 
but not so if it be immersed in any other position. 
For in this case the pressures on equal elements of 
area are not equal, since the pressure varies with 
the depth (§ 24), and the different elements of area 
into which the whole surface may be divided are 
at different depths below the surface of the fluid. 
Consequently, the centre of pressure, which is the 
point in, the plane surface at which the resultant of 



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3» 



Hydrostatics. 



all these forces acts, does not correspond with the 
centre of gravity of the surface. 

The term centre of pressure is used with respect 
to plane surfaces only, since it is not always possible 
to find a single force equivalent to the resultant action 
of a fluid on a curved surface. 

§ 38. To find the centre of pressure of a rectan- 
gpilar area immersed vertically. 

Let A B c D be a rectangle having its upper side 
A B in the surface of the 
liquid Then ifEF be drawn 
to bisect A B and d c, the 
pressure will be equally dis- 
tributed on each side of e f, 
and the centre of pressure 
will lie somewhere in this 
line. 

To •determine where, 
take M N bisected by f to 
represent the pressure at r. Join em, en. Then if 
we take any point p in e f, and draw q p r parallel 
to M F N, Q R will represent the pressure at p. 

For, since the pressure varies with the depth, the 

pressure at p : the pressure at f : : e p : e f 

and ep:ef::qr:mn 

/. the pressure at p : the pressure at f : ; q r : m n 

and /. Q r represents the pressure at p. 

Now the problem of finding the point of action of the 

resultant of a number of forces represented by such 

lines as q r, acting at all points of e f, is the same as 

that of finding the centre of gravity of the triangle e m n. 

But the centre o^ gravity of this triangle is kno\ni to be 




I> M 



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Centre of Pressure, 39 

at a point g in e f, such that f g = J f e, and therefore 
the centre of pressure is at the same point, i,e. at a 
distance of one-third up the middle line from the 
base. 

§ 39. To find the centre of pressure of a triangle 
having one of its sides in the surface of the liquid. 

If the triangle be divided into a number of nar- 
row horizontal stnps, the pressure on each strip acts 
at its middle point, and therefore the whole pressure 
on the triangle acts somewhere in the median line 
drawn through the middle point of the side in the 
surface of the liquid. But the pressure on each strip 
is proportional to its area multiplied by its depth, 
and the value of this product is constant for every 
pair of strips at equal distances from the middle 
point of the median line. Hence the resultant 
pressure acts at the middle point of this line, or the 
depth of the centre of pressure is half that of the apex 
immersed. 

§ 40. To find a general expression for the centre 
of pressure of a plane surface. 

Suppose the surface divided by horizontal lines 
into any number of small elements, then if a be the 
area of one of these elements and z its depth (which 
may be considered the same as the depth of its 
centre of gravity) below the surface of a liquid of 
sp. gr. J, azs is the pressure on that element of 
area. 

Hence, if -s^i, 2^2, . . . . be the depths of the several 
elements, the whole pressure equals 

a\Z\S-\'aiiZi^s-\-a^z^sAr 

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40 



Hydrostatics. 



and if Z be the depth of the centre of pressure, then 
by the principle of parallel forces 

or Z= ^i^i^+^2^^+g3V+. . . . . 

The general application of this method, as of the 
similar method for finding the centre of gravity of a 
surface, requires the use of the integral calculus, but 
special cases may be determined by ordinary alge- 
braical processes. 

§ 41. Bzamples. — (i.) To apply the method of § 40 to 
find the centre of pressure of an isosceles triangle with its 
apex in the surface of the liquid and its base horizontal. 

Suppose the triangle to be divided by horizontal lines into 
narrow horizontal strips. 

Let h — height of triangle, h its base ; and let s be the 
sp. gr. of the liquid. 

Let h be divided into n 
parts, then the breadth of each 




of these strips will be -. 


Also, 


since F E ; E A : : B D 


: DA 


/. F E = — A E, and the area 
2h 



of each strip may be represented 

by - • A E •— - = — A E, and 
2h ft 2n 

therefore the pressure on each strip may be taken as equal to 

- / A E y. ^, supposing the depth of the strip below A to 

correspond with that of its centre of gravity. Now since 

— , we have 
n n n n 



A E has the several values — , ?-» 3 f . 



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Fluid-pressure 09t Surfaces immersed. 41 



z - 






± 
2n 

n 

h_ 
n 

h_ 
4 

h_ 
4 



I^ + 2» + 3^+ 



I«+ 22+32 + 

. 4 

236 

236 



2« 3 6»* 



(by summation of series.) 



and if n be infinite 



each of the quantities — and \ has for its limit zero, 

ft fr 

Hence Z ^ ^ , h. 
4 

Or, the centre of pressure is on the median line, and at a depth 
of three - quarters of the 
height of the triangle, from 
its apex. 

(2.) Required the mag- 
nitude and position of the 
resultant pressure on a flood- 
gate, the level of the water 
being different on either side. 

Let A B be a section of 
the flood-gate, and let the 
height of the water on one 
side be a, and on the other 
side ^, and suppose h greater 
than a. 

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Fia 


19. 






^ 








■^SZ— 


rzz-:^. 












e;:::^ 







^. 













IT" 


z:r:zr:irz:=~- 


Birz: 


7^:zz 










H 


?-s^^ 


^ 


:^ 




-J 


j"^ 





/^2 Hydrostatics, 

Then, if >6 be the width of the gate, the total pressure on 
one side will be — 



2 2 



^p. 



and on the other side — ^^5= Q ; and P acts at a point c such 

2 

that B c=s one-third of Bw, and Q acts at a point D such that 
B D =: one- third B« (§ 38). The resultant of these forces Q-P 

equals (^— a') — , and the point E where it acts can be deter- 
mined by the principle of parallel forces. Thus : 

* ' 2 2 2 

2\ 3 3/ 

• B ^-t±hist 

Exercises. IV. 

1. A cubical block each edge of which is 5 cm. is sunk 
in water, with two opposite faces horizontal ; find the difference 
in the pressures on its lower and upper surfeces. 

2. A cylinder 10 inches high contains liquid to the height 
of 8 inches : find the line of action of the total pressure on the 
interior surface of the cylinder. 

3. A hollow cube is three-fourths filled with water. One of 
the sides of the cube moves freely about a hinge at the base. 
Required the force that must be applied at the upper edge of the 
moveable side and perpendicular to it to keep it in equilibrium. 

4. Find the height of a cylinder the diameter of which is 
2 feet, so that the whole pressure on the curved surface may 
be four times as great as the pressure on the base, when the 
cylinder is filled with liquid. 

5. A rectangle is immersed in water with one side in the 
surface. Show how to divide it by a horizontal line into two 
parts on each of which the whole pressure shall be the same. 

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Fluid-pressure on Surfaces immersed, 43 

6. A rectangular v«sel has a partition 10 inches high and 
8 inches broad. On one side of the partition is water to the 
height of 4 inches : on the other side alcohol (sp. gr. o*8) to the 
height of 6 inches. Find the magnitude and position of the 
resultant pressure on the partition. 

7. An equilateral triangular lamina is immersed in water 
with one side in the surface and the opposite angle 2 deci- 
metres below its surface. If each side of the triangle mea- 
sures 6 decimetres, find the total pressure which the water 
exerts on it. 

8. A rectangle a BCD is immersed in water with the 
side A B in the surface. Find the pressure on each of the tri- 
angles formed by the diagonal a c, and show where it acts. 
Show also that the resultant of these two pressures coincides 
with the pressure on the whole rectangle. 

9. Find the height to which water may rise on one side of 
a wall 2\ metres high and half a metre thick without overthrow- 
ing it, the specific gravity of the material of the wall being 2. 

10. A cylinder, height h inches and diameter 2r, contains 
three liquids of specific gravities .S",, S^ .S",, which do not mix 
in layers of equal thickness : find the whole pressure on the 
base. 



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44 Hydrostatics. 



CHAPTER HI. 

FLUID PRESSURE ON BODIES IMMERSED. 

VII. Resultant Vertical Pressure, — Principle of 
Archimedes, 

§ 42. We have seen (§ 32) that if a body is 
wholly immersed in a fluid, every point of its surface 
is subjected to a pressure perpendicular to the surface 
at that point. 

Now, all these pressures, acting as they do in 
various directions, can be resolved into horizontal and 
vertical components ; and since the horizontal pres- 
sures equilibrate each other, the resultant pressure 
must be vertical, and act upwards or downwards. 
Moreover, since the pressure varies with the depth, it 
is clear, speaking roughly, that the whole pressure on 
the lower half of a body is greater than that on the 
upper half; and, hence, the resultant of all the 
pressures on a body immersed is a force acting verti- 
cally upwards. This force is called the resultant 
vertical pressure, 

§ 43. Experiment. — Take an ordinary weight of 
one pound, and having attached it to a piece of strong 
thread, let it hang wholly immersed in water from one 
of the scale-pans of the hydrostatic balance, as shown 
in fig. 22. It will now be found to weigh less than 

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Fluid-pressure on Bodies immersed, 45 

I lb., and this difference of weight is due to the re- 
sultant vertical pressure acting upwards. 

§ 44. Measure of the Besultant Vertical Pres- 
sure. — If a symmetrical body with vertical sides be 
immersed in a fluid, we can easily see what the mag- 
nitude of this resultant pressure is. For, since the 
downward pressure exerted by the fluid on the body 
is equal to the weight of the column of fluid having a b 
(fig. 20) for a base, whilst the upward pressure is equal 
to the weight of the column having c d for a base, the 
resultant vertical pressure must be equal to the dif- 

FiG. 20. Fig. 21. 



ference between the weight of these two columns, i,e, 
to the weight of a column of fluid equal to a b c d, i,e, 
to the weight of the fluid displaced. 

If the body be of irregular shape, a more general 
method of proof must be employed. Thus : 

Suppose the body immersed to be a portion of the 
fluid itself solidified. Then, if no change of density 
take place, the solidified fluid will remain as before in 
equilibrium. Hence the weight of this portion of the 
fluid, which acts, at its centre of gravity, vertically 
downwards, must be counterbalanced by the upward 
pressure of the fluid ; and consequently the resultant 

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46 Hydrostatics. 

vertical pressure equals the weight of the solidified 
fluid. But the fluid would exert exactly the same 
pressure on any other body occupying the same space 
in the fluid. Hence the resultant vertical pressure on 
any body immersed is equal to the weight of the fluid 
displaced, and acts at its centre of gravity, which point 
is called the centre of displacement^ or centre of 
buoyancy. The principle thus established is commonly 
known as the principle of Archimedes ^"^ and may be thus 
enunciated : 

When a body is immersed in a fluid it is subject 
to a force equal to the weight of the fluid displcued^ 
which acts at the centre of buoyancy vertically upwards. 
Or, A body immersed in a fluid loses a portion of its 
weight equal to the weight of the fluid displaced. 

Before proceeding to consider some of the chief 
deductions from this proposition, we will show how 
it may be experimentally verified. 

§ 45. Experiments. — i. Take a vessel witii a spout 
in one side, as shown in fig. 22. Pour in water till it 
Fig. 22. begins to run out from the spout. 

Take a piece of iron weighing i lb., 
and having suspended it by a fine 
thread from one of the scale-pans 
of the hydrostatic balance, weigh it 
in the water, allowing the water 
displaced to escape into another 
vessel, B. The piece of iron will 
^ be found to weigh nearly 14 oz., 
and the weight of the water col- 
lected in the vessel b will be found to be a little 
* Born at Syracuse in Sicily, flourished about 250 B.C. 

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Principle of A rchimedes, 47 

more than 2 oz., thus showing that the loss of 
weight of the body is equal to the weight of the water 
displaced. 

2. Take a hollow brass cylinder, a, into which 
a solid cylinder, b, exactly fits. Hang the hollow 
cylinder to one of the Fig. 23. 

scale-pans of the hy- 
drostatic balance, and 
attach the solid cylinder 
to it by means of the 
hook, as shown in fig. 23. 
Now weigh carefully the 
two cylinders. Having 
observed their weight, 
take a vessel containing 
water, and let the solid 

cylinder hang in it completely immersed. Equi- 
librium is destroyed, and the free scale-pan de- 
scends. Now fill the hollow cylinder with water, 
and equilibrium is at once restored, clearly showing 
that the weight of the water in the hollow cylinder, 
i,e, the weight of a quantity of water of the same 
size as the body immersed, is equal to the loss of 
weight of the solid in water, Le, to the upward pressure 
which the liquid exerts on the body. 

§ 46. Beal and apparent Weight of Bodies. — 
We have seen that a body surrounded by a fluid is 
pressed upwards by a force equal to the weight of the 
fluid displaced. This is the case with all bodies 
weighed in air, and consequently their real weight 
(i.e. their weight in vacuo) is greater than their ap- 
parent weight by the weight of the air displaced 

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48 Hydrostatics. 

If two bodies, the volumes of which are not equal, 
balance one another in air, their real weights are not 
the same : that which has the greater volume is really 
the heavier. Thus, let the volumes of the two bodies 
be V and V\ their real weights W and W, Then, 
if they balance in air, W^ Vs=: W— Vs, where s is 
the specific gravity of air. 

Hence, W^W' + { F- V% or, 
^is greater than W, if Fis greater than V. 

When the weights of two bodies, of small and 
nearly equal volumes, are compared, the diflference 
between their real and apparent weights is so slight 
as to be in most cases of no practical importance. But 
where the volume of one body is much greater than 
that of the other, the quantity ( F— V')s cannot so 
easily be neglected. Thus, in answer to the question. 
Which is the heavier, a pound of feathers or a pound 
of lead ? we should say that the apparent weights of 
both are the same ; but the real weight of the feathers 
is greater than that of the lead, and the two would not 
equilibrate each other in vacuo. 

§ 47. Bzamples.— (i.) A solid cube of metal the edge of 
which is 3 inches, and whose specific gravity is 7, is wholly 
immersed in water and is supported by a string attached to it. 
Find its apparent weight in water. 

Let Wh^ the weight of the body in air, which differs very 
slightly from its real weight, A its apparent weight in 'wSiter, 
and y the resultant vertical pressure acting upwards. Then the 
body is in equilibrium under the action of these three forces, 
and, therefore, W^ Y+A, 

Now, the resultant vertical pressure K equals the weight of 

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Fluid-pressure on Bodies immersed, 49 

water displaced «3» xze;, where w=» weight of a cubic inch of 
water, and W^ 3* x 7 x w. 

Hence -4« W— y«3*x 7xw-3»xw = 6x3»xz(/ 

162 X 1000 ^ „ ^ - 

° .1728 °'- = Slbs. i3|o^. 

(2.) A body weighs in vacuo 560 grams, and in water 
60 grams : find its volume. 

Here W^ A = 560 - 60 = 500 grams a weight of water dis- 
placed. Hence volume of the water displaced, ue., the volume 
of the body = 500 cubic centimetres. 

{3.) Two hollow spheres, the volumes of which are 100 and 
200 cubic centimetres respectively, balance one another in 
vacuo. What weight must be placed inside the larger that they 
may balance in -water ? 

Weighed in water the larger sphere will seem to be the 
lighter, since the force supporting it is greater. Now the force 
supporting the larger sphere in water is 200 grams, and the 
force supporting the smaller sphere is ic» grams. Hence, for 
equilibrium, the weight of the larger must be increased by 100 
grams without increasing its volume. This may be done by 
placing 100 grams inside. 

{4.) Find the acceleration with which a heavy smooth body 
will sink in a perfect fluid less dense than itself. 

Let ff^-the weight of the body, 5 its absolute specific 
gravity. 

Let / = the specific gravity of the fluid. 

Then the volume of the body is — - (§ 17) and the weight of 

5 

the fluid displaced is W—, 

Hence, the resultant force measured in gravitatic*n units, 
causing the body to descend, is W ( i— -^ and the mass of 

the body moved is — . 



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50 Hydrostatics, 

Ifi therefore, / is the acceleration with which the body de- 
scends, it is shown in books on Mechanics that 

If the body be specifically lighter than the fluid, the body, if 
placed under the fluid, will tend to rise to the surface, and 

the acceleration /= ^-^ . g. 

s 



Exercises. V. 

1. Find the weight in water of a piece of zinc (sp. gr. =7) 
that weighs 8 oz. in vacuo. ^ *" 

2. Find the volume of a body that weighs 350 grams in 
vacuo and 225 in water. ' • u j ^ - T i J - liJ^t V^ 

3. A block of stone 2 cubic feet is wholly immersed in . 
water. With what force is it buoyed up ? ". 3 • i '^ I 1 I Xf^tjb 

4. A piece of metal weighs 36 lbs. in air and 32 lbs. in fresh 
water. What will it weigh in sea-water the sp. gr. of which is 

1-025? y-<jvj-. V -- ^-z 

5. An air-ball, the volume of which is 6 cubic decs., weighs 
in air 15 grams. Find its real weight, having given that the 
weight of I cubic centimetre of air is 0*0013 grams. 

6. A round disc of lead (sp. gr. = 1 1 '35) area 5 sq. centi- 
metres and thickness 2 centimetres, is fixed to a piece of cork 
(sp. gr. =o*24) of same area and 8 centimetres thick. Find the 
weight of both in water. 

7. The edge of a hollow cube of lead is 8 centimetres, its , 
tliickness is 2 centimetres. Find its weight in water. /'•^•*' ^^^ " '^''^ 

8 A bottle weighing 200 grams is completely filled with 
water, when it is found to weigh 800 gr. If a piece of iron 
(sp, gr. =7*2) is placed in the bottle, the bottle with its contents 
weighs 955 grams. Find the weight of the iron. 

9. A piece of metal of specific gravity 8, and weighing 



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Fluid-pressure oft Bodies immersed. 5 1 

20 lbs., is dropped into a cylinder Jilledmth water. Find the 
additional pressure on. the base. 

10. A cylinder of wood (sp. gr. «=o*75), one decimetre high, 
with a sectional area of 16 sq. cm., is immersed with its axis 
vertical in alcohol (sp. gr. =o*8). Find the least weight that 
must be placed on the top of it to bring its upper surface on a 
level with the alcohol. 

11. A piece of wood weighing 8*25 grams (sp. gr. bo*66) is 
placed 4 ft. 7 in. deep in water and is free to rise. N^lecting 
all frictional resistance, find its velocity when it reaches the sur- 
face of the water. 

12. Find the time occupied by a stone (sp. gr. b3*2) in 
falling from rest through 55 feet of water. 

13. Two spheres whose radii are respectively I^ and r cm., 
and both of which are heavier than their respective bulks of 
water, are of equal weight. What weight of metal (sp. gr. =j) 
must be attached to the doUom of the larger that they may 
balance each other in water? 

14. Two masses of given specific gravities balance when 
suspended from the equal arms of a lever in a known fluid. 
What is the specific gravity of a fluid in which they balance 
when one of the masses is doubled ? 

15. A cubic inch of one of t\ro liquids weighs a grains, and 
of the other 6 grains. A body immersed in the first weighs / 
grains and in the second g grains. What is its reai weight and 
what is its volume ? 

VIII. F/oafing Bodies — Metacentre, 

§ 48. When a body is immersed in a fluid, we 
have to consider three separate cases : — 

I. The weight of the body may be greater than 
the weight of the fluid displaced, in which 
case motion will take place in the direction of 
the greater force, and the body if left to itself 
will sink. 

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52 Hydrostatics, 

2. The weight of the body may be equal to the 

weight of the fluid displaced, in which case 
it will rest anywhere in the fluid. 

3. The weight of the body may be less than that 

of the fluid displaced, in which case the re- 
sultant vertical pressure will force the body 
upwards, and it will float 

These three cases can be easily illustrated by ex- 
periments. The first of them has been already con- 
sidered in the preceding paragraphs, and it has been 
shown that in order that the body may be prevented 
from sinking in the fluid, it must be upheld by a force 
A equal to its apparent weight, such that 

when ^is the weight of the body in vacuo, and Y 
the resultant vertical pressure, or weight of the fluid 
displaced. 

The second occurs less frequently, and does not 
now need to be separately treated. Later on, when 
speaking of the formation of drops, we shall have 
occasion to consider the form assumed by a mass of 
fluid which is wholly supported by the external re- 
sultant pressure. 

The third case is of great practical importance, 
and brings us to the consideration of floating bodies. 

§ 49. Principle of Flotation.— If we take a body 
the weight of which is less than the weight of an 
equal volume of a liquid into which it is immersed, 
the resultant vertical pressure will bring it to the sur- 
face, and the body will be found to assume a posi- 

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Principle of Flotation. 53 

tion of equilibrium in which it will be only partly 
immersed. If the experiment indicated in § 45 (i) be 
tried with a body lighter than Fig. 24. 

water, it will be found that the 
weight of the liquid that escapes 
in consequence of the partial im- 
mersion of the body is exactly l^^^^^^^Sf 
equal to the weight of the body ]|OI 
itself. ' ^ 

The relation between the part of the body which 
is immersed and the part that rises above the sur- 
face of the fluid is determined by the principle that 
the two forces acting on the body must equilibrate 
each other ; ue. the weight of the whole body acting 
downwards must equal the weight of the fluid dis- 
placed acting upwards. Thus if a b (fig. 24) be the 
intersection of the body with the surface of the liquid 
in which it floats, the position of a b is determined by 
the equation W =■ Y^ where ^ is the weight of the 
body, and Fthat of the liquid displaced Moreover, 
if G (fig. 24) be the centre of gravity of the body and 
s the centre of buoyancy, g and s must be in the 
same vertical line, for otherwise the body would be 
acted upon by a couple which would produce oscil- 
lation or rotation. Hence the conditipns of a body 
floating in perfect equilibrium are : — 

1. The weight of the body must equal the weight 
of the fluid displaced. 

2. The centres of gravity of the body and of the 
fluid displaced naust He in the same vertical line. 

§ 50. Stable and unstable Equilibrium. — If a 
floating body be slightly displaced, so that the points 

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54 



Hydrostatics. 



s and G do not lie in the same vertical line, the mo- 
ment of the forces acting on the body may either 
tend to restore the body to its original position, as in 
fig. 25, in which case the equilibrium is said to be 
stable^ or to overturn the body altogether, as in fig. 26, 
in which case the equilibrium is unstable. 

Whether a floating body, partly immersed, can 
suflfer a small displacement without being overturned 




is a matter of the greatest practical importance, as the 
safety of all vessels at sea depends on it For, in 
consequence of the action of the tides and waves, a 
vessel seldom or never floats in perfect equiHbrium, 
but is contmually undergoing a rotatory displacement, 
which makes it oscillate about its original position of 
equilibrium Now we see that if s is above G the 
body will alv^ays return to its original position, or in 
this case the equilibrium of the vessel may be said to 
be stable. But where s is below G, the body is very 
likely to overturn if displaced, and hence the danger 
of overloading the deck, and the advantage of ballast 
in lowering the centre of gravity of the vessel. But 
a body may float in stable equilibrium even if s be 
below G, as we shall see presently ; and consequently 

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Stable and Unstable Equilibrium. 55 

we cannot determine from the relative positions of s 
and G only whether the equilibrium of a floating body 
is unstable or not. 

§ 51. Metacentre. — When a body floating in equi- 
librium receives a slight displacement, the centre of 
gravity of the fluid displaced assumes a new position 
on \^hich the stability of equilibrium mainly depends. 
Thus let ABC (fig. 27) be a body floating in equili- 
brium, and let s be the centre of buoyancy. Suppose, 
now, the body to undergo a slight displacement, in 

Fig. 37. 



consequence of which the centre of buoyancy changes 
to s'. If, then, the vertical through s' meets in the 
point M, the line through s and G which was vertical in 
the first position of the body, it is clear that when m is 
above g the equal forces acting on the body will form 
a couple tending to restore the body to its former 
position, and that when m is below G they will tend 
to overturn the body. The point m is called the 
metacentre of the body, and its position depends on 
the shape of the body and on the position of its centre 
of gravity. Of course, if g is below s the point m will 
be above g. Hence it follows generally that the 
equilibrium of a floating body is stable or unstable 

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56 Hydrostatics, 

according as the metacentre is above or below the centre 
of gravity of the body. 

The metacentre may be defined as the point in 
which the vertical through the centre of buoyancy of 
a floating body which has undergone a slight dis- 
placement, intersects the line drawn vertically through 
the centre of buoyancy in the original position of equi- 
librium. 

If a body floats wholly immersed in a fluid, no 
change in the position of the body will alter the 
relative positions of the centres of gravity and buoy- 
ancy. Hence in this case the equilibrium cannot be 
stable unless the centre of gravity is below the centre 
of buoyancy. If the positions of these points be 
reversed, the slightest oscillation will cause the body 
to overturn. 

§ 52. Bxamples.— (I.) A body the specific gravity of the 
material of which is s floats in a liquid specific gravity s'. Find 
what fraction of its volume will be immersed. 

Nearly all problems on floating bodies may be solved by the 
application of the principle of Archimedes, i.e, by equating the 
weight of the floating body with the weight of the fluid it dis- 
places. 

Let Fequal volume of the body; V of the portion immersed. 

Then K^ = weight of body; and F'/s: weight of fluid dis- 
placed ; 

V s 

(2.) A cylinder the height of which is 12 inches floats two- 
thirds immersed in water. Find what part of it will remain im- 
mersed if a liquid the specific gravity of which is o*2 be poured 
upon the surface of the water so as to completely cover the 
cylinder. 

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Fluid-pressure on Bodies immersed. 57 



When the cylinder floats in the water only, the weight of 
the cylinder equals the weight of the water displaced. 

where s is the specific gravity and a the sectional area of the 
cylinder. 

,*, specific gravity of cylinder = §. 

When the lighter liquid is poured on to the surface of the 
water, the weight of the cylinder equals the sum of the weights 
of the fluids displaced. 

If z equals in this case the part of the cylinder immersed in 
the lighter liquid we have 

I2xf = 2X0*2+(I2 — 2) 

or 80 = 22+ 120- IQ2 
.', 2 a. 5 inches. 

Hence the cylinder rises one inch out of the water in conse- 
quence of the additional upward pressure due to the liquid 
poured into the vessel. 

(3.) Required the weight that must be attached to the end of 
a straight rod of known weight and specific gravity, that it may 
float vertically in water. 

A uniform straight rod cannot float in a vertical position, 
because the centre of gravity is always above the centre of 
buoyancy. If, however, a heavy Fig. 28. 

particle of no considerable magni- 
tude be attached to the lower end 
of the rod, the centre of gravity may 
be lowered as much as we please. 
What is required, therefore, is the 
weight of a particle that will lower 
the centre of gravity to the depth, at 
least, of the centre of buoyancy. 

Let G be the centre of gravity 
of the rod A b, a its sectional area, and s its absolute specific 
gravity, and lV\\s weight. Let x be the weight attached to the 
end A, and g be the centre of gravity of ^and x. Then 




w+a> 



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58 



Hydrostatics, 



{IV+ x)kg=: JVx A G where ^= 2AGxaxs 

Also, by principle of Archimedes, IV-k- x^AK,a 

a 
Now, in order that the rod may float vertically, Ag must no 
be greater than A s. Put A^« A s. Then 
H^ _ W+x 
{IV+x)2as 2a 

.'. ^^{iv+xy 

This is the /easi weight that will suffice. Any greater weight 
than this may be added to A, provided it is not great enough 
to completely immerse the rod. Hence, the maximum weight 



possible is x, where — = IV +x, 



or x^fV 



(7-)- 



The weight required, therefore, must have a value some- 
where between 



^(^_,)a„d^(i-,) 



In the above calculation the volume of the heavy particle, being 
P small, has been altogether neglected. 

(4. ) A uniform rod of given length 
and specific gravity moves freely about a 
point at a given height above the sur- 
face of a liquid in which it is partly 
immersed. Required its position of 
equilibrium, 

I>et A B be the rod ; PV, Y, G and s 
as before. 

The position of equilibrium is deter- 




"^^- 



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Fluid-pressure on Bodies immersed. 59 

mined by equating the moments of W and Y about A. Tak- 
ing the sectional area, which is uniform, as unity we have 

ff^=ABxjand y=BKx^, 
where s and d are the specific gravities of the material of the 
rod and of the liquid respectively. Also 

^xGM=yxSN /. if=^JL 

Y GM 

. -iLuL^Af by ^imilaiity of triangles. 
BK. rf AG "^ ' ^ 

• AB S _ A B + AK _^AB 

' ' AB-AK * d 2 * 2 

•■•?-'-(^)' 

•••rfvo-j) •••— 'vo-a- 

If A c is given, A K can be expressed in terms of A c and 
of the angle B a c, which is thus determined. 

Exercises. VI. 

1. A cylinder of wood (sp. gr. =0*6) and 12 inches high 
floats with its axis vertical in water. To what depth will it be 
immersed ? 

2. A cube of oak (sp. gr. =097) each edge of which is 
6 inches, floats partly in sea water (sp. gr. i '028) and partly in 
olive oil (sp. gr. =0-915). Find what part of it is immersed in 
each liquid. 

3. A uniform block of metal 10 inches high (sp. gr. = 8) floats 
in mercury (sp. gr 13*6). Plnd how much it rises out of the 
mercury if water be poured on the surface of the mercury so as 
to completely cover the block of metal. 

4. A cylinder floats vertically in a liquid. Compare the 
forces necessary to raise it and to depress it to an equal extent. 

5. A heavy uniform rod weighing 3 kils. and sp. gr. = 6 
moves freely under water in a vertical plane about a hinge at one 
end. If a string tied to the other end supports the rod in a 
horizontal position, find the tension in the string. 

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6o Hydrostatics. 

6. A piece of gold (sp. gr. 19*25) weighing 96*25 grams 
immersed in a vessel full of water causes 6 grams of water to 
be displaced. Is the gold solid ? If not, find the size of the 
hollow. 

7. Two equal globes, the volume of each being 100 c.c, are 
suspended from the equal arms of a lever, the one hanging com- 
pletely immersed in water, the other in a liquid of sp. g^. =0*8. 
What additional weight is required to make them balance ? 

8. A cube of metal is floating in mercury (sp. gr. = 13*6). 
When a weight of 170 lbs is placed on the top, it is observed to 
sink 3 inches. Find the size of the cube. 

9. If an iceberg (sp. gr.= 0*918) float in sea-water, what 
is the ratio of the part submerged to that which is seen above 
water ? 

10. Find what quantity of cork must be attached to a man 
whose weight is 168 lbs. and sp. gr. = i -12 so as to enable him 
just to float in water. 

11. An iron ball of 12 lbs. weight floats in mercury covered 
in water. Find the weights of the parts in the two fluids ; having 
given specific gravity of mercury =13*6, specific gravity of iron 

= 7-5. 

12. The specific gravities of the upper and lower of two 
fluids that do not mix are 0*9 and i*i ; the- upper fluid is 4 in. 
deep ; a cube with an edge of i foot and sp. gr. 0*75 floats in 
the liquids ; how much of it is immersed ? 



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specific Gravity, 61 



CHAPTER IV, 

SPECIFIC GRAVITY, AND MODES OF DETERMINING IT. 

IX. Application of the Principle of Archimedes to the 
determination of the Specific Gravity of Bodies, 

§ 53. To find the specific gravity of a substance we 

require to know its weight in vacuo, and its volume, 

W 
since 7^ = ^f- The volume of the body, expressed m 

cubic centimetres, is numerically equal to the weight 
of an equal bulk of water at the standard tempera- 
ture expressed in grams, since the weight of a unit- 
volume of water is one gram ; and, therefore, the 
specific gravity of a substance is expressed numerically 
by the ratio of its weight to that of an equal bulk of 
water. 

Instead of the weight of the body in vacuo, we 
generally substitute the weight in air, the difference 
being unimportant for solid and liquid bodies of small 
magnitude. Where the volume of water equal to that 
of the body can be directly found, the specific gravity 
is very easily determined ; but where this is not the 
case different methods have to be employed. 

The principle of Archimedfts tells us that a body 
immersed in a fluid is pressed upwards bv a force 

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62 Hydrostatic!:. 

equal to the weight of the fluid displaced. Hence the 
specific gravity of a body can be found by comparing 
the real weight of a body with the difference between 
its real weight and its apparent weight in water, since 
this difference is equal to the weight of water dis- 
placed. If, therefore, W represent its real weight, 
and A its apparent weight in water, the specific 

W 
gravity of the body equals ^ . 

§ 54. To find fhe specific gravity of a solid body 
insoluble in water. 

This can be determined roughly by an experiment 
similar to that described in § 45 (i), where the weight 
of the body may be directly compared with the 
weight of the water displaced. But a more accurate 
result may be obtained by using the hydrostatic 
balance for finding A^ the weight of the body in 
water. The specific gravity can then be determined 
by dividing W^ the weight of the body, by W^A^ the 
loss of weight in water ; or 

specific gravity = -^ — - 

% 55. To find the specific gravity of a solid body 
that fioats in water. 

In this case the body may be said to have a nega- 
tive weight in water : Le, it requires the application of 
a positive force to make it weigh zero when completely 
immersed. Let ^ equal the weight of the body, P 
the force required to completely immerse it, then the 
force necessary to resist the resultant vertical pressure 
= ^ + P = weight of water displaced by the whole 
body. 

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specific Gravity. 63 

W 



Hence, specific gravity = 



W^-P' 



What we want, therefore, is to determine P, This 
we can do by attaching a heavy body called a sinker 
to the light body, and by weighing them both in 
water. If they weigh zero, the sinker's weight in 
water is the force P required ; but if they tend to 
sink and are found to weigh -5, then the sinker's 
weight in water is too great by this weight B. 

If, therefore, A equals the apparent weight of the 
sinker in water, 

A-^B^P, 

and the specific gravity of the body is r^jz :; =. 

fr + A — o 

For determining these weights the hydrostatic balance 

is again employed. 

§ 56. To find the specific gravity of a liquid, by 
weighing a solid in it. 

Let the solid weigh W in air, and let a be its ap- 
parent weight in the given liquid. 

Then ^—dJ = weight of the volume of liquid 
displaced by the solid. 

Let solid weigh A in water. 

Then W— A^ weight of the volume of water 
displaced by the solid. 

But these two volumes are the same ; 

.•, specific gravity of liquid = " 

§ 57. To find the specific gravity of a solid body 
soluble in water, but insoluble in a liquid of known 
specific gravity. 

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64 Hydrostatics, 

If the body is soluble in water, its weight in water 
cannot be directly found, and consequently the 
method of § 54 is inapplicable to this case. If, 
however, the body is insoluble in some other liquid 
the specific gravity of which is known, we have the 
means of determining the specific gravity of the 
body. 

Suppose the weight of the body to be W^ and its 
weight in the known liquid a^ then W — a \% the 
weight of the liquid it displaces ; and if s be the spe- 
cific gravity of this liquid, — ^^ = the volume of 

• s 

liquid occupied by the body. It follows, therefore, 

W—a 
that the weight of an equal volume of water is , 

since specific gravity of water is unity, ^ 

W—a TVs 
Hence, specific gravit}' of body= JV^ — 



W-a 

§ 58. Specific Oravity of Oases. — In determining 
the relative weight of gases, air at 0° C, and at the 
ordinary atmospheric pressure is taken as the standard 
substance. The process is attended with so many 
difficulties, owing to the peculiar properties of gases 
which have not yet been considered in this work, that 
many precautions are needed in order to obtain accurate 
results. In comparing the weights of equal volumes 
of any gas and air, it is necessary that the gases should 
be subjected to the same atmospheric pressure, and 
should be brought to the same temperatiure. 

If we suppose JV to be the weight of a glass globe 
full of air, and w its weight when empty, and if fV' 

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specific Gravity, 65 

be the weight of the same globe filled with any other 
gas, then the specific gravity of this gas as compared 
with air is 

W— w 

If the globe be filled with water from which all 
air-bubbles have been carefully excluded, and if Wx 
be its weight so filled, the specific gravity of the gas 

as compared with water would be -— . 

^ Wx—w 

In determining with accuracy the specific gravity 
of any substance, the temperature at which the experi- 
ment is conducted must be considered; for it is 
known that bodies generally expand when heated, 
and the weight of the same volume of water, or of any 
other substance, consequently varies at different tem- 
peratiures. This subject is more fully considered 
in treatises on heat 

Before working the following exercises the student 
is recommended to take pieces of brass, tin, zinc, 
marble, wood, and other easily obtained substances, 
and to determine their specific gravities by the hydro- 
static balance, comparing the results with those given 
in the tables. 

Exercises. VII. 

1. A solid soluble in water but not in alcohol weighs 346 
grams in air and 210 in alcohol. Find the specific gravity of 
the solid, that of alcohol being 0-85. 

2. A solid weighs 100 grams in vacuo, 85 grams in water, 
and 88 grams in another fluid. What is the specific gravity of 
the fluid? 

F 

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66 Hydrostatics. 

3. A piece of copper weighs 31 grains in air and 27*5 in 
water. Find its specific gravity. 

4. A piece of wood (sp. gr. ==074) of 32 cubic inclies floats 
in water. How much water will it displace ? 

5. A body floats in water with one-eighth of its volume above 
the surface : determine its specific gravity. How much of it 
will be submerged in a fluid whose specific gravity is 09 ? 

6. A body floats in a liquid (sp. gr. =si3*5), ^^^%. ^^ ^^s 
volume is above the surface : find specific gravity of the body. 

7. If a ball of platinum weigh 21*4 oz. in air, 20*4 oz. in 
water, and 19-6 in hydric-sulphate, find the specific gravity of 
the platinum and of the acid. 

8. A piece of marble, specific gravity = 2*84, weighs 92 grams 
in water and 98*5 grams in oil of turpentine. Find the specific 
gravity of the oil. 

9. A piece of cork weighs 2 oz., and its specific gravity is 
0-24. What is the least force that will just immerse it? 

10. A piece of iron, sp. gr. 7*21, and weighing 360*5 grams 
is tied to a piece of wood weighing 300 grams, and the weight 
of both in water is 110*5 grams. Find the specific gravity 
of the wood. 

11. A cubic block of wood each edge of which is 8 cm. is 
put into water. If the specific gravity of the wood is 0*85, with 
how much more wood must it be loaded, so that its upper sur- 
face may sink to the level of the water ? 

12. A cylinder of wood 20 inches in height, and a cylinder of 
lead I inch in height, are united so as to form one cylinder 
21 inches in height, which is found to float in water with 
3 inches projecting above the surface. Find the specific gravity 
of the wood, that of the lead being 1 1 -4. 



X. Ot^er Methods of obtaining the Specific Gravity of 
Substances. — Hydrometers. 

§ 59. The Specific Gravity Bottle.— This is a 
bottle made to hold a certain weight of water at the 



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The Specific Gravity Bottle. 6^ 

standard temperature. It may be employed for deter- 
mining the specific gravity of a liquid or a powder. 

1. Specific Gravity of a Liquid, — Suppose the bottle 
when empty to weigh w grams, and that it holds 
I GO grams of water. Let the bottle be filled with 
the liquid the specific gravity of which is required, 
and suppose it then to weigh (91+ a/) grams. Then 
the weight of the liquid in the bottle is 91 grams, 
and the weight of an equal bulk of water is 100 
grams. 

Hence, specific gravity of liquid is -^ = 0*9 1. 

2. Specific Gravity of a Powder, — Let the powder 
be first placed in the bottle, and let the bottle be then 
filled with water, and suppose the weight of the 
contents of the bottle, /.^., of the powder and water, 
to be K, Then if ^be the real weight of the powder, 
and y the weight (unknown) of the water it displaces, 

^^=100+ w- y, 
and .-. Y^W-^-ioo-K 

W 
/. Specific gravity of powder = -j^ 

Take 4 grams of powdered glass, put it into 

the bottle, and fill with water ; then the bottle with 

its contents will be found to weigh 102*5 + 0/ grams ; 

/• F=4— 2-s=i-5; .•, specific gravity=Ttj=2-6. 

§ 60. Hydrometers.— The hydrometer^ consists 

essentially of a straight stem, loaded at one end, so as 

* This instrament is said to have been invented by Hypatia, 
the daughter of Theon Alexandrinus, who flourished about the 
end of the fourth century ; though there is some foundation for 
the opinion that tht invention is due to Archimedes. 
F 2 

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68 Hydrostatics, 

to float in a vertical position. By observing the 
depths to which it sinks in two different Hquids, the 
relative weights of these two liquids can be deter- 
mined. Hydrometers are of very different forms, and 
may be used for finding the specific gravity of liquids 
or of solid bodies. We shall describe two varieties 
only. 

§ 6i. Common Hydrometer. — ^This instrument, 
invented by Fahrenheit,* consists of a straight stem, 
Fig. 3a usually made of glass, which terminates in two 
^ hollow spheres. The lower sphere is loaded 
^ with mercury, so that the instrument may 
float in a vertical position. 

Let a = section of the stem. 
w = weight of the instrument. 
V = volume „ „ 

Suppose the instrument to float in water with * 
the point d of its stem on the surface of the 
water, and in some other liquid, the specific 
gravity of which is to be found, with the point c on the 
surface. 

Then, if j be the specific gravity of the liquid, it 
follows from the principle of Archimedes that 

'w = si^—a X A c) = v—a x a d 
taking the weight of the unit-volume of water as 
unity. 

• __ Z^~gXAD 

§ 62. Nicholson's Hydrometer.' — This form of 
instrument enables us to determine the specific gravity 

> 1724. * 1787. 

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Hydrometers, 69 

of solid as well as of liquid bodies. It consists of a 
hollow body, b, connected by a fine stem with a small 
dish, A, at its upper end, and at its lower end with a 
cup, c, loaded so as to ensure equilibrium. 

When floating in distilled water, with a certain 
weight, k^ in the upper dish, the instrument sinks 
to such a depth that a fixed mark, 0^ is on the surface 
of the water. 

1. If we want to find the specific gravity of a given 
liquid, we place the instrument in it, 

and place weights in the upper dish ^^^' 3'' 
till the fixed mark, ^, is on a level with 
the surface of the liquid. 

If, then, W be the weight of the 
instrument, k the weight added to sink 
it to in water, and w be the weight 
added to sink it to in the given liquid, 
W-^-k is the weight of the water dis- 
placed ; and W-^-w is the weight of 
the liquid displaced ; and, as the same 
bulk of fluid is displaced in each case, 
the specific gravity of the liquid is 

W+k' 

2. To find the specific gravity of a solid. 

Let k be the weight as before that must be 
placed on the instrument to bring the point ^ on a 
level with the surface of the water. Place a small 
piece of the solid on a, and diminish the weight k^ so 
as to keep the instrument at the same level. 

Let m be the weight now employed. 

Then k—m=si weight of solid body. 

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yo Hydrostatics. 

Now place the substance the specific gravity of 
which is to be found in the cup c, in which case a 
weight greater than m by the weight of the water 
displaced by the body must be placed on a to main- 
tain the instrument in the same position. 

Let n equal the weight now in a. 
Then «—f« = the resultant vertical pressure on the 
solid in c, 
= the weight of the water displaced by it 

Hdnce, specific gravity is -— ^. 

Exercises. VIII. 

1. A piece of glass the weight of which is 50 grains is placed 
on the upper dish of a Nicholson's hydrometer, and it is found 
that an additional weight of 235 grains is required to sink the 
instrument to the fixed level. If, however, the glass be placed 
in the lower cup a weight of 250 grains is required. Determine 
the specific gravity of the glass. 

2. A globe of glass holds 5 litres of water at standard tem- 
perature. When full of water it weighs 5500 grams. "When 
full of air it weighs 506 "465 grams, and when full of hydrogen 
it weighs 500*447 grams. Find the specific gravity of air 
as compared with water, and of hydrogen as compared with air. 

3. A bottle filled with water is found to weigh 500 grams. 
If 180 grams of powder are introduced into the bottle, the 
bottle with its contents weighs 575 grams. Required the 
specific gravity of the powder. 

4. Into a specific gravity bottle capable of holding looo 
grains of water, 300 grains of a certain powder are introduced. 
The bottle is then filled up with a liquid, specific gravity o*8, 
and the contents of the bottle are found to weigh 980 grains. 
Find the specific gravity of the powddr. 



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The Motion of Liquids. 7 1 



CHAPTER V. 

THE MOTION OF LIQUIDS. 

XI. Liquids moving by their own weight, 

§ 63. It is proposed to consider in this chapter a 
few of the most elementary propositions connected 
with the movements of fluid bodies. Very little can 
be attempted without the application of some of the 
higher processes of mathematics, and we shall endea- 
vour, therefore, only to indicate the principles of the 
methods on which the solution of this class of pro- 
blems depends. As all liquids are more or less viscous, 
absolute agreement cannot be expected to exist 
between the conclusions arrived at on the assumption 
of a frictionless fluid and the results of direct ex- 
periments. 

§ 64. DefinitioD. — The vertical distance between 
the surface level of a liquid in a vessel and the centre of 
the orifice through which it escapes is called the head 
ox pressure height under which the flow takes place. 

§ 65. Torricelli's Theorem. — The velocity with 
which a liquid escapes from a small hole in the bottom 
or side of a vessel was the subject of numerous experi- 
ments made by TorricellL^ The result at which he 

* Born in Italy in 1608 ; died in 1647. 

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72 Hydrostatics. 

arrived was that the rate of efflux was the same as 
would be acquired by a body falling freely from the 
surface level of the liquid to the centre of the orifice 
from which it escapes. 

Let A B c D be a vessel having a small open- 
ing, o, in one of its sides. Let the area of this open- 
Fig. 32. ing, which is supposed to be very 
small, be ^, and suppose a small mass 
of liquid, ;;/, the thickness of which is 
Cy to occupy this opening. 

Then if / be the pressure in excess 

of the atmospheric pressure urging 

this mass forwards, measured in units 

of force per unit-area, p=^g. da h 

where d is the density of the liquid, and h the pressure 

height, o H ; and the work done by this force p in 

urging the mass m from a position of rest through 

the distance c is 

p X c =i g, dah.c) 

and since the work done is equal to the energy which 
the mass m acquires, 

p , c '=' , where v is velocity of efflux ; 



sr.d ahc=i dac — 
^ 2 

or v^ -=■ 2 g h 

/>., the velocity is that which would be acquired by 
the mass m in falling freely from the surface level of 
the liquid to the centre of the orifice. 



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TorricellVs Theorem, 73 

In this result each particle of the mass is supposed 
to move at right angles to the area of the orifice, and 
to escape into the air at a pressure equal to that at the 
surface level 

§ 66. Experimental Test.— The accuracy of the 
results given by Torricelli's theorem may be tested 
for different liquids, by fig. 33. 

taking a vessel having 
apertures in one of its 
sides through any of 
which the water or other 
liquid it contains may 
escape. If the vessel 
be kept constantly full, 
and one of these aper- 
tures be opened, the « 6 c 
liquid will flow through 

it, and will descend in a curve similar to the path of 
a body projected horizontally from a certain height 
above the ground. 

If the distances payjf>b,pche observed, the accu- 
racy of Torricelli's theorem can be tested. 

For, let the height o c = ^, then according to the 
theorem, if v be the velocity of the liquid at c, 

V = -^ 2gh, and the velocity will be uniform and 
horizontal. 

If then / equal the time occupied by a particle in 
moving from c to c, we know by the second law of 
motion that / equals time occupied by a body falling 
freely from c to / 

= time of describing/^ with the uniform velocity v. 

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74 Hydrostatics. 

Hence, pc^='tv\ also cp = — ^ 

orvj=pcx^/ S ^ 

V 2C/ 

Now, if the distance/ c be known, and if /^ be ob- 
served, the value of z/ so found can be compared with 
that given by the formula z/ = n/ (2^. o c). On 
comparing the results, obtained by experiment, with 
those obtained from Torricelli's theorem, a certain 
difference will be always found to exist. Some of 
the reasons of this difference, apart from those depen- 
dent on the viscosity of the liquid, will be considered 
later on. 

§ 67. Belation of Telocity of Flow to Sectioiial 
Area of Vessel* — Let a b c d be a vessel, which is 

Fig. 34. 




kept constantly full of liquid. We may consider the 
mass of the fluid, which is supposed to be incom- 
pressible, to be divided into small laminse moving 

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Vena Contracta. 75 

parallel to one another and of uniform sectional area 
for very small differences of depth. Thus, libcde, 
p^rsyhe two small laminae, we may suppose bc^=^de 
and/^ ^^ r s. Let each particle in the section be 
have the velocity, Vy and each particle in the section 
pq the velocity v'. If, then, the section be descend 
to d e in one second, the section / q will descend 
to some depth, rs, in the same time ; and the volume 
bcde will be equal to the volume /^rj, since the 
quantity of liquid between de and /^ remains con- 
stant 

If, then, A equal the section be ox de, and a' the 
section/^ or r J, 

A z/ = a' z/' = volume discharged in one second, 
/. V \ v' \\ a! : A, 
or the velocities are inversely proportional to the 
sectional areas. 

§ 68. Vena Contracta.— When a liquid issues 
into the air from a small opening in a thin plate, the 
stream is found, first of all, to converge, so that it 
contracts rapidly for some little distance from the 
orifice. This fact may be very easily observed, and will 
be seen in all cases such as those shown in figs. 33, 34. 
The area of the jet at its narrowest part is known 
as the vena contracta^ and was first made the subject of 
investigation by Sir Isaac Newton. It is found to 
be generally about three-fifths of the aperture of the 
orifice, but varies with the shape of the aperture and 
the pressure height. The value of this fraction, which 
is called the eoefficient of eonir action ^ can be determined 
by experiments only. In all calculations with respect 
to the rate of the emptying of vessels, the method 

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jS Hydrostatics. 

usually adopted has been to apply the results of Tor- 
ricelli's theorem, and to substitute the area of the vena 
contracta for that of the orifice. This method, we 
shall see, gives us a merely empirical result, and is 
faulty in so far as it overlooks some important elements 
in the problem. 

§ 69. MeasTire of Pressure on the Walls of a 
Pipe. — Suppose a liquid is flowing through a pipe 
A B, the pressure which it exerts on the walls of the pipe 
at any point can be experimentally determined by in- 
serting gauge glasses, />., thin glass tubes of about 

Fig. 35. 



f of an inch diameter, into the pipe at those points 
at which the pressure is to be found. Since fluids 
transmit their pressure equally in all directions, the 
forward pressure exerted by the moving liquid will be 
equally exerted on the walls of the pipe, and will 
force the liquid up the gauge glasses to a height corre- 
sponding to the pressure produced. If the pipe is of 
uniform sectional area the pressure produced at all 
points will be the same, friction being neglected, and 
the liquid in the pipe will rise in the gauge glasses 
to the same height at all points in the pipe. 

§ 70. Eolation of Pressure to Sectional Area of 
Pipe. — If the sectional area of the pipe varies, the 

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The Motion of Liquids through Pipes. Jj 

pressure is no longer uniform, but is found to be 
greatest where the area is greatest, and vice versd. 

Suppose the liquid to be moving with a uniform 
velocity in that part of the pipe a b (fig. 35) which 
has a uniform sectional area. If, then, we consider an 
element a of the liquid, we see that the pressure on 
either side of a must be the same, since any increase 
of pressure from behind would cause it to move with 
an acceleration, which is not supposed to be the 
case. 

Suppose now the element to have reached b, and 
to be entering the wider part of the pipe, its velocity 
in the direction of the pipe's length will now be less 
than it was before, and will continue to decrease as 
the pipe widens. Hence the pressure of the liquid 
behind the element a urging it on must ever be less 
than that in front of it resisting its advance, and con- 
sequently the pressure must increase with the area 
of the pipe and with the decrease in the velocity of 
motion. 

After passing c, the widest part of the pipe, the 
velocity begins again to increase, and as the resist- 
ance to the forward motion of the particles must 
consequently have become less, the pressure is also 
diminished. 

These results can be practically illustrated by 
observing the height of the liquid in the several gauge 
glasses placed at different points in the pipe. Due 
allowance must be made for the frictional resistance 
which causes the liquid to rise to a less height in the 
glasses than it otherwise would, and gives a difference 
between the theoretical and actual results which in- 

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7S 



Hydrostatics. 



creases with the length of pipe traversed. This re- 
sistance being neglected, we see that the pressure 
which a liquid exerts on the sides of a pipe through 
which it is passing varies directly with the sectional 
area, the velocity at the two ends of the pipe being 
the same. 

If/i and/z be the pressure exerted at points where 
the area of the pipe is a^ and azy we have 

A :/2::«i :«2- 

§ 71. Belation of Velocity of Flow to the Pres- 
sure produced. — We have seen that in the case of a 
steady flow through a pipe of varying area, the velocity 
at different points varies inversely with the area, fric- 
tion being neglected. We have now to consider how 
the pressure changes with the velocity. This may be 
experimentally exhibited by fixing the larger end of a 
tapering pipe into a vessel of water kept constantly 
full, the upper surface having only the ordinary at- 
mospheric pressure upon it. If the pipe be furnished 

FiQ. 36. 




with gauge glasses, the pressure at different points 
can be observed. The hydrostatic pressure at a is 
evidently zero, or the same as that of the air into 

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The Motion of Liquids. 79 

which it issues, but the liquid will be found to stand 
at a continually higher level in the gauge glasses as 
the pipe widens towards the vessel. The difference 
between the surface level of the water in any one of 
the gauge glasses and that of the vessel is called the 
fall of free level dX that point, and this fall is equal to the 
difference between the height of the column of liquid 
representing the hydrostatic pressure at a point in the 
stream, and the pressure-height at the same point, 
supposing the liquid to be in equilibrium. 

Consider now a small mass m of the liquid in the 
section of the stream at any point b, and let the 
section of this mass be equal to a unit of area, and 
its length equal to ^, so that it contains c units of 
volume. Then if/ be the actual pressure in units of 
force per unit of area at b, and if P be the pressure 
at the same point due to the depth of the mass m 
below the free surface l l' d of the liquid, then the 
whole amount of work which the mass m has received 
from the pressure behind it in moving through its 
own length from a position of rest is -Px r, and since 
/ is the pressure it is exerting, the work which it gives 
to the liquid in front of it is/ x c. Hence the excess 
of work which it receives is {P—p) c. 

Now if JjTis the vertical depth of m below l l', and 
h the height of the liquid in the gauge glass above niy 

P=:gHdzxidp — gdh, 
where d is the density of the liquid, 

.-. {P^p)c^gd{H^h)c. 
But the amount of work gained by the mask m \^ 
equal to its energy. 

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8o Hydrostatics. 

Hence, g d^H—h) c = ^^, where v is the velocity 

of mass m. 

And since m = dCy we have 

Z;2 = 2g {H—h) ^ 2gDG 

or, the velocity at any point in the stream is that due 
to the difference between the statical and actual pres> 
sure-height of the moving particles. Hence, m steady 
flow the velocity generated from rest is that due to the 
fall of free level. 

If we suppose the sectional area at b to be twice 
that at A, and at c twice that at b, and so on, then if 

V be the velocity at a, the velocity at b = ^, since the 

velocity varies inversely with the area in cases of 
steady flow \ 

and since z/^ = 2 ^ . a d, 

we have — =• 2^.dg, orBG = ^AD; 

4 4 ' ' 

and — = 2 ^. D H, or c h=— ^ a d, and so 

16 16 

on, which gives the relation between the actual hydro- 
static pressure at a point, and the velocity with which 
the stream is passing that point 

§ 72. Application to the Flow of a Liquid through 
an Orifice in the Base or Side of a Vessel.— When a 
liquid flows through an orifice, as previously con- 
sidered in the case of Torricelli's theorem, the flow 
does not actually take place in a direction at right 

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Flcnv of Water through Small Orifice, 8i 

angles to the orifice, at all points of the orifice, but 
va directions such as are shown in fig. 37. The 
nature of the flow is therefore 
^'^' 37' somewhat similar to that indi- 

cated in the preceding section, 
in which the varying area of 
the pipe is represented by the 
space included between the 
several stream lines. At all 
points along the margin of 
the orifice the direction of the 
motion is tangential to the 
plane of the orifice, and the water comes at once into 
contact with the atmosphere. But at all intermediate 
points the issuing stream exerts a pressure greater than 
that due to the atmosphere, and consequently the 
velocity at any of these points is not that due to the 
height of the free surface of the liquid in the vessel, 
but is that due to the difference between this height 
and that of the column of liquid corresponding to 
the actual pressure at the particular point in the 
moving stream. This difference of surface level 
has already (§ 71) been referred to as the fall of 
free level ; hence the velocity of any particle in 
the issuing stream is that due to the fall of the 
free level \ or, if h equals the depth of a particle in 
the stream below the free surface of the liquid, and 
z equals the height of the column of liquid represent- 
ing the pressure at that particular point, then if v be 
the velocity with which this particle is moving, 
(I = s/zglh—z). 

In calculating the actual discharge in a given time 

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82 Hydrostatics. 

through a small orifice in the base or side of a vessel, 
the following principles, which are essential conditions 
of the flow, have to be considered : 

1. The absolute velocity at any point in the plane 
of the orifice is not that due to the vertical distance 
between the point and the free surface level of the 
liquid, since everywhere throughout the area, except 
along the boundary of the orifice, the liquid is 
under a pressure greater than that of the atmos- 
phere, and consequently, as we have shown above, 
the velocity is correspondingly diminished, or 
V s= \/ 2 g(/i — z), 

2. If the orifice be divided into small horizontal 
bands, we cannot suppose the velocity to be constant 
throughout each band, since the direction of the 
motion of the fluid is different at different parts. 

3. If we consider any vertical section of the liquid 
as shown in the figure, since the direction of the mo- 
tion towards the orifice is nowhere perpendicular td 
the orifice, except at its centre, the actual velocity of 
efflux is only the component of the whole velocity 
as determined by the first of these principles, acting 
in a direction at right angles to the plane of the orifice. 

These considerations show how very complicated 
is the problem of determining the actual discharge o 
a liquid, per unit of time, through a given orifice, and 
serve to indicate some of the causes of the discrepan- 
cies between the results of actual experiments and 
those given by Torricelli's theorem. Ordinarily, when 
the orifice is small, the discharge is roughly calculated 
by considering the velocity of efflux to be that due to 
the distance between the centre of the orifice and the 

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Flow of Water through Small Orifice. 83 

level of the free surface, and by then substituting the 
area of the vena contracta for that of the orifice. Thus 
if a be area of a very small orifice, and y the co- 
efficient of contraction, and if h be the vertical dis- 
tance of the centre of the orifice from the free sur- 
face level, the discharge per unit of time is taken as 
y a ^/2 g h. But this result is not only a mere 
empirical result, the degree of accuracy of which de- 
pends on the value of y as experimentally determined, 
but it leaves out of consideration some important cir- 
cumstances under which the flow actually takes place. ^ 
§ 73. Effect of Friction on the Pressure of a 
Liquid Flowing Vniformly througli a Pipe.— Hitherto 
we have taken no account of the frictional resistance 
which retards the passage of a liquid through a pipe, 
and which, apart from all other circumstances, causes 
the velocity of efflux to be less than that due to the 
entire pressure-height above the orifice from which 
the liquid escapes into the air. 

The effect of this frictional resistance on the pres- 
sure exerted by a liquid in flowing through a pipe of 
uniform sectional area may be conveniently illustrated 
by the following arrangement. 

H A (fig. 38) is the section of a cylindrical zinc vessel 
about 3 ft. high, into the side of which, near the bottom, 
is fixed a brass pipe of uniform bore. At equal dis- 
tances along this pipe are openings into which glass 
gauge-glasses are fitted. 

If the opening e be closed, and the apparatus 
filled with water, the liquid will stand at the same 

> *0n the Flow of Water through Orifices.* By Prof. 
James Thomson. Report of British Association, Glasgow, 1 8 761 

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84 Hydrostatics. 

level in the vessel and in the pressure-tubes ; but as 
soon as the end e of the tube is opened, the water 
falls to different levels in the several tubes, being 
lowest in the tube nearest the end, and rising by equal 
differences in the several tubes. If the water in the 
vessel be kept at the same level, it will be observed 

Fig. 38. 



that the surface levels of the liquid in the several tubes 
lie on a straight line drawn through e to some point e'. 
This line is found to be more nearly horizontal, the 
greater the sectional area of the pipe through which 
the water flows. 

Now it is evident that if the liquid diiring its flow 
had encountered no frictional resistance, its velocity 
on leaving the vessel a h would have been that due to 
the pressure-height above it, and its pressure on the 
tube would not have exceeded that of the atmosphere 
into which it issued. The additional pressure on the 
tube is due, therefore, to the frictional resistance 
encountered by the liquid ; and as the resistance to 
be overcome at any point of the tube increases with 
the distance of that point from the orifice, the pressure 
of the liquid diminishes as it approaches the open end. 
Let a equal the sectional area of the pipe, and let c be 

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Effect of Friction. 8 5 

the length of an element of liquid occupying it Let/ 
be the measure of the frictional resistance per unit of 
length of pipe, and let p and/' be the actual pressures 
per unit area at either end of the element c. Then if 
we suppose m to be the mass of this element, and %f 

its velocity, which is constant, — represents the 

2 

energy of this mass at all points in the pipe. 

Now if the mass m moves through a distance r, the 
work which it has received \^ pa x ^, and the work 
which it has given to the liquid in front of it \^p' a x c. 
Hence the resultant work gained in moving through 
the space c\^(p a — p' a) c. But since the energy of the 
mass remains constant, this amount of work is em- 
ployed in overcoming the frictional resistance encoun- 
tered by the liquid in its passage through the pipe. 
The work done against friction is/ x c\ and 

.% (p—p')ac^fy.x\ 
and since / — Z' ^=^gd(h—h') 

where h^ h' are the heights of water required to pro- 
duce the pressures / and ^ respectively, and d is 
density of the liquid, we have 

gda 
which shows that the difference of pressures (as mea- 
sured by height of the liquid in pressure-tubes) for 
equal distances along a pipe of uniform sectional area 
is constant. 

It is to be here observed that whilst the kinetic 
energy of the flowing water is constant, there is a 
continual decrease of potential energy much in the 
same way as when a body slides with a uniform 
velocity down a rough inclined plane* zed by Google 



86 Hydrostatics, 

XII. Capillarity. 

§ 74. In the preceding sections we have treated of 
the equilibrium and motion of liquids in mass, but we 
shall now consider a number of phenomena which are 
mainly due to the action of forces on the molecules 
of a liquid, and which are not immediately explicable 
by the principles already stated These will be dis- 
cussed under the two heads of Capillarity and Diffusion, 
and they comprise the formation of drops, the relation 
of liquids and solids in contact, and the intermixture 
of liquids, either in contact with one another, or 
separated by porous membranes. In discussing these 
subjects we shall in each case commence with a few 
experiments, and then indicate the nature of the prin- 
ciples which seem to explain them. 

§ 75. Drop Formation. — ^Experiments.— Take a 
vessel containing olive oil and carefully drop into it 
some small portions of water. These will be seen to 
assume a spherical shape as they descend slowly 
through the oil Raindrops are spherical; but the 
rapidity with which they fall through the air is so 
great that we are unable to see them long enough to 
distinguish their form. 

To show that a large amount of liquid will assume 
a spherical shape if left free to the action of its inter- 
nal forces, mix together alcohol and water, in such 
proportions that the mixture may have tho density of 
olive oil This will require about three-parts of alco- 
hol to one of water. Then pour gently, so that it 
may not be broken into parts, some olive oil into the 

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Formation of Drops, 87 

mixture, and it will be found to float in the form of a 
sphere in any part of the surrounding liquid. With^ 
out any great difficulty globules of oil may in this way 
be obtained, having a diameter of four or five inches. 
These can be flattened between two glass plates, or 
receive an indentation by being touched with a glass 
rod, without losing their cohesion ; and as soon as the 
external body is removed, they recover like an elastic 
ball, their spherical form. 

If pure quicksilver be scattered on a level surface 
of glass, the smallest of the drops will differ very little 
in form from perfect spheres. The larger drops, owing 
to their weight, will be considerably flattened ; and if 
a greater quantity cohere together, the tendency to a 
spherical shape is visible, only in the somewhat curved 
form of the upper surface. 

In blowing an ordinary soap-bubble we obtain a 
very good idea of the nature of the forces that act on a 
fluid surface. For this purpose * take some common 
soapsuds, or a Plateau's mixture of soap and glycerine, 
and blow a small bubble at the end of a tube with a 
bell mouth. A tobacco-pipe will answer if the bore 
of the tube is large enough. After blowing the bubble 
at one end of the tube, place the other end near the 
flame of a candle. The bubble will contract and 
drive a current of air through the tube, as may be seen 
by its effect on the flame. 

*This shows that the bubble presses on the air 
within it, and is like an elastic bag. To enlarge the 
bubble by blowing into the tube, work must be done 
to force the air in, because the pressure inside the 
bubble is greater than that of the air outside. This 

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88 Hydrostatics, 

work is stored up in the film of the soapsud, for it is 
able to blow the air out again with equal force.' 

§ 76. Surface-Tenrion.— -These experiments show 
that when two fluids are in contact with each other 
and do not mix, the surface separating them is in a 
state of tension similar to that of a membrane 
stretched equally in all directions, and that the 
surface-particles have a tendency to approach one 
another like those of the outer side of a bent watch- 
spring. This contractile force, or surface-tension, 
resides in the thin film separating the two fluids, and 
does not extend to any appreciable depth below the 
surface ; and to its action is due the spherical form 
of the drop and of the soap-bubble. In the case of 
the soap-bubble the superficial tension may be mea- 
sured by considering the work done in order to pro- 
duce a film of a certain area \ and the measure of the 
tension per unit length is found to be the same as the 
numerical value of the work done divided by the area.* 

This tension seems to be due to the fact that 
whilst the particles in the interior of a fluid are sub- 
jected to equal forces on all sides, those on the sur- 
face separating the two fluids are under the influence 
of different molecular attractions, the result of which 
is to give to the bounding surface a tendency to con- 
tract. The surface-tension depends on the nature 
and temperature of both media, decreasing as the 
temperature rises, and vanishing altogether at that 
temperature at which there is no longer a well- 
marked distinction between the liquid and gaseous 
states.^ Between any two liquids that do not mix, 

> C. Maxwell, Theory of Heat, pp. 281, 283. - § 4. 

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Surface-tension, 89 

between a liquid and its own or another vapour, 
and between a solid and a fluid of any kind, there 
is a definite surface-tension, the value of whigh 
for a unit of length is called the co-efficient of superfi- 
cial tension, or co-efficient of capillarity, as having been 
first considered in connection with the ascent of 
liquids in capillary tubes. This tension may be 
reckoned in c. g. s. units of force per linear centimetre, 
and it is found that at a temperature of 2o°C. the 
tension of a surface of water in contact with air is 
81, that of mercury and air 540, that of alcohol 25*5, 
and of olive oil 36'9 units of force. In fact, of ordi- 
nary liquids water is found to have the greatest sur- 
face-tension. 

§ 77. Capillary Elevation and Depression. — 
Experiments. — (i.) If a sheet of clean glass be im- 
mersed in water and then removed, the water will be 
found to have wet the surface of the glass. If, whilst 
partially immersed, the surface of the water in contact 
with the glass be observed, it will be found that the 
surface of the water in the immediate neighbourhood 
of the sheet of glass is raised. If we perform the 
same experiment with mercury instead of water, we 
„ shall find that the mercury 

Fig. 39. Fig. 4a , , , -^ 

does not wet the glass, 
and that the surface of 
the mercury in the neigh- 
bourhood of the glass will 
be depressed. 

If A B be the surface 
of the liquid, c d the vertical section of the plate of 
glass, the water (fig. 39) will be found to rise to some 

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90 Hydrostatics, 

height a b^ and the mercury (fig. 40) to be depressed 
to the level a' b\ Moreover, the surfece of the water 
in contact with the glass will be concave, and that of 
the mercury convex \ and the elevation of the water 
will be found to be somewhat greater than the 
depression of the mercury. 

(2.) Take now two plates of glass and let them ap- 
proach each other in water. When they are suffi- 
ciently close, the water will begin to rise between them, 

Fig. 41. 



and the height to which it rises will increase as the 
distance between them lessens. Moreover, the sur- 
face of water between the plates will be concave. If 
the same experiment be made with mercury instead of 
water, the mercury will be depressed between the 
plates, but to a less extent, and the surface will be 
convex. 

If this experiment be varied by taking plates of 
different thickness, the same results will be obtained, 
thereby showing that whilst the amount of ascent or 
depression depends on the distance between the plates, 
it is independent of their thickness. 

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Capillary Phenomena. 



91 



Fig. 42. 




(3.) If instead of plain surfeces parallel to one 
another we take two plates 
of glass inclined at an angle, 
and immerse them partly in 
water, the water will rise be- 
tween the plates to different 
heights, and the intersection 
of the surface of the water 
with the plate will form a 
curve, which is known as the 
rectangular hyperbola. 

(4.) If glass tubes of small bore, called capillary 
tubes (from capiila, a hair), be partially immersed in 
water, the liquid will rise in the tubes, and the height to 
which it reaches will be found to vary inversely with 
the diameter of the tube. This result holds good 
whether the experiment be made in air or in a vacuum, 
but the elevation is foimd to diminish as the tempera- 
ture of the water increases. If mercury be used 
instead of water, the liquid will be depressed, as pre- 
viously stated in the case of parallel plates. The 
curved surface of the liquid is called a meniscus^ and 
the meniscus is concave for water and convex for 
mercury. 

§ 78. Principles of Capillarity.— When a liquid 
comes in contact with a solid, the particles of the 
liquid on the common surface of the two substances 
are to some extent in a similar condition to those 
of the free surface of a liquid ; />., they are not 
equally attracted on all sides, as is the case with 
particles in the interior. The interaction of the 
molecular forces of the surface of the liquid and solid 



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92 Hydrostatics, 

gives rise to a tension in the common surface that 
separates them. When a solid is in contact with two 
fluids there is a different tension in each of the 
surfaces separating a pair of media. In this case, 
in order that the three tensions acting at any point 
in the Hne of intersection of the surfaces may be in 
equilibrium, their directions must be such that they 
can be represented by the sides of a triangle, and 
any one must be greater than the difference between 
the other two. Now we have seen that the surface 
of a liquid in contact with a solid is curved, and 
for any given liquid in contact with a given solid 
there is a definite angle of contact, called the angle of 
capillarity^ which is cunite in the case of mercury, and 
obtuse in the case of water in contact with glass. This 
inclination of the surface of a liquid to that of a solid 
at the margin of contact is due to the excess of the 
tension of the surface separating the two fluids (for 
example, air and water, or air and mercury) above 
the difference of the tensions of the surfaces separating 
the solid from each of the fluids. 

§ 79. Height of ElevatioiL or DepressioiL of 
Liquid in Capillary Tubes. — Let T denote the mag- 
nitude of the superficial tension per unit length of 
the free surface of the liquid inclined at a given angle 
to the solid in contact with it. Then the vertical 
component of T acting upwards or downwards is 
the force tending to raise or depress the liquid. 
Let / be the vertical component of T', and let r be 
the radius of the tube, then 2 xr/ denotes the magni- 
tude of the vertical force acting all round the tube. 
If, now, s be the specific gravity of the liquid, h the 
mean height to which it is raised or depressed, then 

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Law of Diameters. 93 

we have nt^hs equal to the weight of liquid supported, 
and 27rr/=7rr^^j, 

ox h = — 
rs 

which gives the law of diameters ^ viz., that the height 
of elevation or depression varies inversely with the 
radius of the tube. 

Since the surface-tension acts uniformly through- 
out the film, it produces a resultant pressure normal 
to the surface, which is always directed towards the 
centre of curvature of the surface. 

It follows, therefore, that in capillary elevation the 
liquid surface must be concave, and in capillary de- 
pression convex. 

In the case of a narrow tube partly immersed in a 
liquid which is in contact with air on both sides of the 
tube, we see that when the surface of the liquid within 
the tube is concave, the pressure immediately below the 
surface is less than the atmospheric pressure by a force 
due to the concavity of the surface, and that the pres- 
sure goes on increasing till at a depth corresponding to 
the mean level of the external hquid it is equal to the 
atmospheric pressure. If the surface is convex, the 
pressure immediately below the surface of the liquid in 
the tube is greater than the atmospheric pressure by 
a force corresponding to the difference of the mean 
level of the liquid inside and outside of the tube. 
Thus, in the case of a concave meniscus the resultant 
normal force due to the surface-tension acts iii a 
direction opposite to gravity, and so lessens its effect, 
and in the convex meniscus it increases it. 

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94 Hydrostatics. 



Xlli. Diffusion of Liquids, 

§ 80. DiffasioiL — If two liquids of different den- 
sities, and susceptible of permanent admixture, be 
placed in contact with each other, they gradually 
become intermixed. The process by which these 
liquids combine is known as diffusion, 

§ 8t. Oraliam's Experiments. — The phenomena 
of diffusion were first carefully investigated by Pro- 
fessor Graham, and the results of his early experiments 
were published in 1850. In conducting these experi- 
ments Graham employed a number of small phials, 
each holding about 114 cc of water. The necks 
were carefully ground and of a uniform aperture of 
about 3*15 c. in diameter. Into these 
'^* ^^' ^ phials he poured solutions of various salts, 
so that the liquid rose as far as the shoulder 
of each phial. The phials were then very 
carefully filled with cold water. Thus 
charged, each phial was closed by a 
glass plate and then placed in a glass 
jar, containing sufficient water to cover 
the mouth of the phial and to rise at least 
2*5 c above it. The plate was then 
carefully removed and the whole apparatus main- 
tained for a certain period of time at a fixed tempe- 
rature. The phial was then withdrawn and the water 
in the jar having been evaporated, the amount of salt 
that had passed from the phial into the jar by this 
process of diffusion was determined by weight. 

A very simple experiment showing the diffusion 



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Diffusion of Liquids, 95 

of liquids is the following : — ^Take a tall cylindrical 
jar and fill it to about two-thirds of its fig. 44. 
height with a solution of litmus. Then^ 
by means of a funnel pour in very care- 
fully some hydric-sulphate, so as to occupy 
the lower part of the jar. If the jar be 
set aside, the acid will be found, after 
the lapse of two or three days, to have 
diffused into the litmus solution, as shown 
by the consequent red colour it will have 
acquired. 

§ 82. Besnlts of Graham's Experi- 
ments. — By varying the experiments with '\ 
respect to the nature of the salt, the den- 
sity and temperature of the solution, and the time 
occupied in diffusion, Graham arrived at the following 
results : 

1. The increase in density in the diffusion product 
corresponds very nearly with the proportion of salt in 
the solution. 

Taking four different solutions containing one, 
two, three, and four parts of common salt to 100 
parts of water, it was found that the quantities diffused 
at the end of equal times were very nearly in the pro- 
portion of I : 2 : 3 : 4, the variation from this result 
not exceeding one per cent. 

2. The quantity of salt diffused in equal times in- 
creases with the temperature. 

From an experiment with a 4 per cent, solution, it 
was found that with a rise of temperature of i5°C., the 
quantity of salt diffused increased somewhat more than 
one-third 

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96 Hydrostatics. 

3. The rate of diffusion for weak solutions of the 
same salt is nearly uniform, but varies considerably 
with the nature of the substance diffused. 

Of the several substances employed in these 
experiments hydric-chloride was found to be the 
most diffusible, and albumen one of the least diffusible. 
Intermediate between these was common salt Com- 
paring common salt with sugar-candy and albumen, 
it was found that with solution of 20 parts of the solid 
substance in 100 parts of water, exposed for eight 
days at a temperature of 16° C, the specific gravity of 
the diffused solutions were respectively i'i26 ; 1*070 ; 
and 1-053. 

The following table shows the approximate times 
of diffusion of equal weights of different substances, 
taking that of hydric-chloride as unity : 

Hydric-Chloride. . . . .1 
Sodic- Chloride . . . •2*33 

Cane Sugar. 7 

Magnesic Sulphate. ... 7 

Albumen 49 

Caramel 98 

4. If a solution be taken of two different salts 
which do not chemically combine, it is found that each 
follows its own rate of diffusion. Thus the inequality 
of diffusion of two different salts supplies a method 
for their separation, to a certain extent, from each 
other. If a mixed solution of two corresponding 
salts of potash and soda be placed in the phial, the 
potash salt being more diffusive than the soda salt 

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Graham's Experiments, 9/ 

will escape into the water outside, whilst the soda salt 
will be relatively concentrated in the phial. 

5. It is also found, in the case of very weak solu- 
tions, that a salt will diffuse into water which already 
contains some other salt in solution, showing that the 
diffusion of one salt is not sensibly resisted by the pre- 
sence of another. 

§ 2^^, CrystalloidB and Colloids. — By consider 
ing the different diffusibiHty of different substances, 
Graham found that all bodies might be referred 
to two classes, which he called crystalloids and col- 
loids. The properties of these two classes of sub- 
stances are very different. Bodies susceptible ol 
crystallisation belong to the former class. They are 
highly sapid ; they generally form a solution which is 
very slightly viscous, and they diffuse rapidly through 
water, or through a porous diaphragm. Colloids, on the 
other hand (so-called from KoXXiy, glue), to which class 
belong starch, gum, hydrated alumina, albumen, gela- 
tine, and many organic compounds, are insipid and 
have very feeble chemical relations. They diffuse very 
slowly in water ; but they form a medium, like water, 
which arrests the passage of other colloids, but through 
which crystalloid substances are capable of diffus- 
ing. A peculiar property of these substances is their 
mutability. They pass very easily from the liquid to 
the curdled condition. They are often largely soluble 
in water, but are held in solution by a very feeble 
force. They never crystallise, and are generally dis- 
tinguished by the sluggishness of the molecules com- 
posing them. 

§ 84. Dialysis. — Many insoluble colloids are per- 

H 

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98 Hydrostatics. 

meable, as water is, to highly diffusive substances, but 
effectually resist the passage through them of other 
colloid substances which may be in solution. If, 
therefore, a solution containing a crystalloid and col- 
loid substance be placed in a vessel containing water 
from which it is separated by a septum, or membrane, 
formed of some insoluble colloid, the crystalloid will 
pass through the septum, and the other less diffusive 
substance will remain behind. This process of sepa- 
ration by diffusion through a colloid septum is termed 
dialysis. 

Experiment, — ^Take a sheet of very thin and well- 
sized paper having no porosity (so that if wetted on 
one side the other side remains dry) ; let the paper be 
thoroughly wetted and then laid on the surface of some 
water contained in a small basin of less diameter than 
the width of the paper. Having made a small de- 
pression in the paper, so as to form a cavity, place 
within it a mixed solution of cane sugar and gum arabic 
containing about 5 per cent of each. In the course 
of twenty-four hours the water below will be found to 
contain about three-fourths of the sugar, which will 
have passed through the paper, leaving the gum behind. 

What takes place during the process of dialysis 
may be thus explained : — A soluble crystalloid is ca- 
pable of separating the water from the colloid, with 
which it is feebly united in the septum. It thus ob- 
tains a liquid medium for diffusion. The soluble 
colloid, on the other hand, is unable to.separate the 
liquid from the colloidal substance of the septum, 
with which it is chemically combined, and is thus 
unable to find a way for its own passage by diffusion. 

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Dialysis — Osmose. 



99 



The process of dialysis is extensively employed for 
separating poisonous crystalloids from organic sub- 
stances with which they may be mixed ; and thus 
separated they yield more easily to the methods of 
chemical analysis. 

§ 85. Osmose. — Intimately connected with the 
phenomena of diffusion is the interchange of liquids 
through porous diaphragms, which is known as osmose, 
Dutrochet ^ was the first to give his attention to this 
subject The endosmometer, an instrument invented 
by him, illustrates this process. It consists of a long 
tube, connected at its lower end with a membranous 
bag which forms a reservoir. This is filled with a 
solution of sugar or gum, and Fig. 45. 

is kept in water. After a time 
the liquid rises in the tube> 
and the level of the water falls 
in which the endosmometer is 
placed. Moreover, traces of 
the substance contained in the 
bag are found in the water out- 
side. This shows that an inter- 
change of the liquids has taken 
place, and that more liquid has 
passed inwards through the mem- 
brane than has escaped outwards from the reservoir. 
To this flowing-in of the liquid Dutrochet gave the 
name of endosmosis^ and he called the reverse process, 
by which the liquid passes out from the membranous 
bag, exosmosis. 

From numerous experiments Graham was led to 

' Nouvelles Recherches sur Tendosmose et Texosmose, 1828. 

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ICX) Hydrostatics. 

infer that in order to induce osmotic action between 
two liquids they must each be capable of acting che- 
mically, but in different degrees, on the diaphragm 
separating them. Where porous materials are used 
not susceptible of chemical decomposition the osmo- 
tic action is very slight When either of the liquids, 
by means of capillarity or by its action on the septum, 
or possibly by both processes, has effected a passage 
through the diaphragm, diffusion takes place, and the 
liquid rises in the tube. No perfectly satisfactory ex- 
planation, however, of these opposing currents of fluid 
has as yet been given. 

The absorption of liquids by the spongioles of 
plants, and the interchange of liquids, which is con- 
stantly taking place in the animal body, in the pro- 
cesses of nutrition and secretion, are probably due to 
osmotic action and liquid diffusion. 

§ 86. Holecnlar Structure of Liquids. — ^The 
phenomena of diffusion are interesting as affording us 
some insight into the molecular structure of liquid 
bodies. For, seeing that the molecules of one liquid 
are readily displaced by the molecules of another 
liquid into which they are diffused, it is evident that 
these molecules must be in a state of constant agita- 
tion and must be capable of continually changing 
their positions relatively to one another. In this way, 
therefore, the interchange of the molecules of two 
liquids in contact with each other can be explained ; 
and the degree of diffusibility of any soluble substance 
would depend on the extent of the excursion which 
the molecules of the solution undergo, or on the dis- 

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Diffusion of Liquids loi 

tance through which a molecule can travel before it 
has its direction changed by impact upon another. 

Thus the sluggishness of a liquid colloid would be 
accounted for by the very small range of the motion 
of its molecules \ and this explanation is supported 
by the fact already referred to, that these liquids are 
frequently highly viscous, and pass very readily into 
the solid state — a condition of matter in which the 
particles, though probably capable of revolving about 
an axis and of oscillating within certain limits about 
their mean positions, are not supposed to undergo any 
motion of translation whatever. 



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t02 Pneumatics. 



CHAPTER VI. 

THE PRINCIPLES OF PNEUMATICS. 

XIV. General Properties of Gases. — Atmospheric 
Pressure, 

§ 87. Pneumatics. — The application of the prin- 
ciples of dynamics to the investigation of the pheno- 
mena presented by gaseous bodies constitutes a branch 
of Hydrodynamics known as Pneumatics. 

§ 88. Expansibility and Compressibility of 
Chtses. Experiments. — The characteristic properties 
of gaseous as distinguished from liquid bodies are 
shown by the following experiments : — 

1. If a small quantity of gas be admitted into an 
empty vessel it will immediately expand, so as to oc- 
cupy the whole of it. 

2. If a bladder containing gas be placed under the 
receiver of an air-pump, from which the air is gradu- 
ally removed, the gas contained in the skin is found 
to expand with the diminution of the external pres- 
sure. 

3. If a cylindrical vessel fitted with a piston con- 
tain a certain quantity of gas, and the piston be 
pressed downwards by a weight or some other force, 

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A tmospheric Pressure, 1 03 

the volume of the gas will be diminished ; and if the 
pressing force be removed the gas will expand. 

It thus appears that the volume, which a certain 
quantity of gas occupies, depends on the pressure to 
which it is subjected, and changes with the size of the 
vessel containing it 

§ 89. The Air. — ^The earth is surrounded by a 
gaseous envelope reaching to a considerable height 
above the earth's surface. It consists of a mixture 
of two gases, nitrogen and oxygen, with small and 
variable proportions of other gases, especially of car- 
bonic-dioxide. By the experiments of MM. Pictet 
and Cailletet, previously referred to (§ 4), air and 
its two principal constituents have been separately 
reduced to the liquid state. Carbonic- dioxide has 
'long since been known to be a liquefiable gas. 

As the majority of experiments connected with 
gases must necessarily be performed in air, it is very 
desirable, at starting, to consider some of its proper- 
ties, as well as the various effects due to its actioa 

§ 90. The Air has Weight— That the air is a 
heavy fluid was not known till comparatively modem 
times. Its invisibility, and its relative position with 
respect to all ponderables underlying it, may have 
helped to conceal this important fact from the know- 
ledge of early observers. It appears that Aristotle,^ 
suspecting the truth of this fact, and wishing to verify 
his belief, weighed a skin first empty and then inflated 
with air \ and finding it to weigh the same in both cases, 
concluded that au: was a weightless fluid. The failure 
of this experiment was due to his having overlooked 

» B.C. 384-322. 

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t04 



Pneumatics. 



Fig. 46. 



the fact that the skin when inflated with air cx:cupied 
a correspondingly larger volume, and that its weight 
was diminished by that of the air which it displaced, 
which was exactly equal to the air which it contained. 
Aristotle's experiment was regarded as decisive for 
many centuries ; and it was not till the time of Gali- 
leo ^ that the air was known to be a 
heavy fluid. It was reserved, how- 
ever, for later philosphers to see in 
this fact the true cause of a variety 
of phenomena which were previously 
unexplained. 

Otto Guericke^ is said to have 
devised the following experiment for 
showing that the air has weight : — 

By means of the air-pump, of 
which he was the inventor, he ex- 
hausted a glass globe, fitted with a 
stopcock, of its contained air. He 
then very carefully weighed the globe, and as soon 
as the scalebeam was perfectly horizontal he opened 
the stopcock. The air immediately rushed in and the 
globe descended. From this he very justly inferred 
that the additional weight which had to be placed in the 
other scalepan to restore equilibrium was equal to the 
veight of the quantity of air admitted into the globe. 
§ 91. Measure of Atmospheric Pressure. Torri- 
cellfs Experiment. — To Torricelli is due a most im- 
portant experiment, which not only shows us that the 
air has weight, but also gives us a measure of its pres- 
sure-intensity at any point. 
• Born at Pisa 1564 ; died 1642. '^ Of Magdeburg ; 1602-1686. 

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TorricelWs Experiment, 105 

Take a glass tube about 34 inches in length, open 
at one end and closed at the other. Fill it carefully 
with mercury, and, placing the thumb over the open 
end, invert the tube with this end under the yiq.^i, 
surface of some mercury contained in a cup. 
On removing the thumb the mercury will be 
found to sink somewhat, and after a time will 
remain stationary in the tube, with its surface- 
level about 30 inches above the surface-level of 
the mercury in the cup. 

Now, since the pressure-intensity is the 
same at all points in the same horizontal plane 
of a liquid in equilibrium, the pressure-intensity 
at any point of the external surface of the mer- 
cury is equal to that along c d. But the pres- 
sure-intensity on c D is equal to h s, where /i is 
the difference of level of the mercury inside and out- 
side the tube, and s is the weight of a unit volume of 
mercury. Hence the intensity of the atmospheric 
pressure on the external surface of the mercury is ^ j* ; 
and the pressure on any area of the same size as the 
section of the tube is equal to the weight of a column 
of mercury having that sectional area for base and 
the difference of level of the surface-level of the liquid 
inside and outside the tube for height. 

The space above the surface of the mercury in the 
tube is called the Torricellian vacuum. It is really 
occupied by mercury vapour, the effect of which on 
the height of the column may generally be disre- 
garded. 

§ 92. EflFect of Atmospheric Pressure. — The 
downward pressure exerted by the atmosphere is the 

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io6 Pneumatics. 

real cause of a number of phenomena which, at one 
time, were explained on the supposition that * Nature 
abhors a vacuum.' 

It was observed that whenever a fluid was forcibly 
removed from a vessel, air or some other fluid took its 
place; and this tendency to occupy a vacant space was 
foimd to be sufficiently strong to counteract gravity and 
Fig. 48. other forces. Thus if the bulb of 

a glass vessel, the stem of which is 
under water, be slowly heated, bub- 
bles of air will rise through the 
water, owing to the expansion of 
the air in the bulb; and if the source 
of heat be afterwards removed, the volume of the air 
will gradually diminish, and the water will rise in the 
stem. The same effect is produced in a tube open at 
both ends by ordinary suction. 

From observations such as these it was inferred 
that Nature abhors a vacuum ; and this general pro- 
position, which expresses, though somewhat vaguely, 
the results of experiments made, within certain 
limits and under particular conditions, was sup- 
posed for many centuries to constitute an undeniable 
law of Nature. Galileo was the first to show that 
this so-called law was not universally true. He 
found that water would not rise in an empty tube 
to a greater height than about 2,Z feet, and he very 
rightly concluded that the force supporting this column 
of water was the pressure of the atmosphere. Nature's 
abhorrence of a vacuum was thus proved to have a 
practical limit Galileo does not, however," appear 
to have recognised all the consequences of his own 

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Barometers, 



107 



discovery. It was left to Pascal, by an extension of 
Torricelli's experiment, to give to this important prin- 
ciple its full development. By performing Torricelli's 
experiment at different elevations above the sea-level, 
Pascal successfully disproved the old theory, which he 
showed was the result of insufficient and circumscribed 
observations ; and by clearly establishing the fact that 
the column of liquid supported varied with the height 
of the atmosphere above it, he proved conclusively 
that the atmospheric pressure was the true cause of all 
the phenomena in question. 

§ 93. Barometers. — ^A barometer in its simplest form 
consists of a tube such as that used in Torricelli's ex- 
periment, and containing a column of liquid supported 
by the atmospheric pressure which it serves to meiasure. 
The height of the column is the difference in level 
between the surface of the liquid in the tube and in the 
cup. This height varies with the liquid employed and 
the conditions of the atmosphere. fig. 49. 

The great specific gravity of mercury 
renders that liquid best adapted for 
the construction of a barometer, as 
the height of the column is corre- 
spondingly small. With a mer- 
cury barometer the average height 
is 76 cm. 

A form of instrument very com- 
monly in use is that known as the 
siphon barometer. 

It consists of a bent tube open 
at one end and closed at the other. Mercury having 
been introduced into the tube in sufficient quantity to 

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lo$ Pneumatics, 

fill the longer leg, the tube is placed in a vertical posi- 
tion when the mercury assumes a position of equili- 
brium, as shown in fig. 49. 

Its principle is the same as that of the instrument 
already described. The pressure-intensity at d, due 
to the column of mercury e d, is equal to that at c, 
due to the column of air above it 

In this form of instrument the cup containing mer- 
cury is not required; the difference of level of the sur- 
faces of the mercury in the two branches of the tube 
measures the pressure due to the atmosphere. 

§ 94. Barometric Corrections. — Absolute pressure 
per unit area. — ^When the barometer is used for deter- 
mining differences of atmospheric pressure at different 
places, or at the same place at different times, certain 
corrections must be made in the observed height of 
the column, of which the following are the principal: — 

I. Correction for Temperature, — A column of 
mercury that measures 76 cm. at 0° C. is found to 
have increased in length when measured at some 
higher temperature, in consequence of the expansion 
of the liquid under the influence of heat. It is neces- 
sary, therefore, in comparing barometric heights at dif- 
ferent places and at different temperatures to calculate 
what the height of the column at each place would 
have been if the temperature had been the same; and 
it is usual to reduce the observed height of the column 
in each case to the height of a column that would 
produce the same pressure at 0° C. In order to ob- 
tain accurate results, a further correction must be made 
for the expansion of the scale on which the measure- 
ments are marked. 

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Barometric Corrections. 109 

2. Correction for Capillarity. — ^When the mercury 
IS contained in a narrow tube, the internal diameter of 
which is less than three-quarters of an inch, the column 
is, as we have seen, sensibly depressed in consequence 
of the tension of the convex surface of the mercury. 
In this case the barometric reading, as reckoned from 
the top of the convex meniscus to the surface-level of 
the mercury in contact with the air, indicates a pres- 
sure a little less than that of the atmosphere, and the 
necessary correction must be added. The error due 
to capillarity is only observable in narrow tubes, and 
is slightly different when the mercury is rising from 
what it is when the mercury is falling in the tube. 

3. Correction for Difference of Sea-level, — Owing to 
the compressibility of air its density at places near the 
sea-level is greater than at places higher up. The 
exact law according to which the density of the air 
decreases as we ascend will be considered later on ; 
but even if the density of the atmosphere were uni- 
form, the pressure-intensity would be greater at places 
of low than of high elevation. In fact, if a barometer 
is carried up a mountain it indicates a continuously 
decreasing atmospheric pressure ; and, from the dif- 
ference in height of the barometric column at two 
places, the difference of the elevation of the two sta- 
tions above the sea-level can be determined. 

But the height of the barometric column is liable 
to frequent fluctuations, owing to occasional and acci- 
dental causes, the effect of which it is often required 
to ascertain. If, with this object, observations of the 
barometer are made at two stations at different heights, 
an addition must be made to the observed reading at 

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I TO Pneumatics. 

each place, so as to reduce it to what it would be at 
the sea-level. Thus, if the barometric column falls, 
on the average, 60 mm. in being carried from the sea- 
level to station A, and 85 mm. to station B, 60 mm. 
and 85 mm. must be added to the observed reading 
at each of those stations. 

4. Correction for the unequal value of g. — In esti- 
mating the pressure of the air on a given area, by the 
weight of a column of mercury, we obtain a result 
which varies with the value of ^, and affords, therefore, 
no uniform standard for the comparison of the atmos- 
pheric pressure in places situated in different latitudes. 
At Paris, where ^= 980*94 cm., the pressure repre- 
sented by a mercury column of the same height is less 
than at Greenwich, where ^ = 981 -i 7 cm. If h equals 
the height of the barometric column, and //the density 
of the liquid employed, the measure of the pressure 
of the air per unit-area x^ghd^ absolute units of force. 
If we put ^ = 76 cm., and//= 13*596, the density of 
mercury, then the absolute pressure at Greenwich per 
square centimetre is 
98117 X76X i3'596=ioi38x 10^ cos units of force. 

It would be practically convenient to take the round 
number 10^ units of force per square centimetre as 
the standard of atmospheric pressure. This pressure 
would be represented at Greenwich by 74*964 cm., 01 
29*514 inches of mercury. 

The standard atmospheric pressure is commonly 
called * an atmosphere ' ; and a pressure twice as great 
is known as ' two atmospheres,' and so on. Thus, a 
pressure of 10^ units of force would be a pressure of 
ten atmospheres. 

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The Siphon, 1 1 1 

§95. Examplei.— (I.) Requiredthe height of a water baro- 
meter when the mercury barometer stands at 30 inches, the den- 
sity of mercury being 13*596. 

Let h be height of water barometer, 
A' „ ,, mercury barometer, 

and let d and d' be densities of water and mercury respectively. 
Then, atmospheric pressure per unit area ^ghd^gh'd! and 
.', .4 = 30 X 13-596 = 34 ft. nearly, 
since ^= i and g is constant. 

(2.) If a mercury barometer standing at 76 cm. be im- 
mersed 4 metres below the surface of a lake, find the height of 
the column. 

A pressure of 4 metres of water is equivalent to a pressure 
of 4 -^ 13*596 metres of mercury, t.^., 29*4 cm. nearly. Hence 
the increase of pressure will be denoted by a rise of 29*4 cm., 
or the height of barometric column will be 105 '4 cm. 

(3.) What is the absolute pressure, in units of force, due to a 
height of 100 metres of sea-water of density i '027, g being 
981? 

The pressure equals 981 x 1*027 x 100 x 100= 1*0075 ^ 'o' 
units of force per sq. cm., and is equal to about 10 atmo- 
spheres. 

J^ § 96. The Siphon. — This is an instrument the 
action of which depends on the atmospheric pressure. 
It is used for drawing off liquid from one vessel to 
another, and consists of a bent tube with arras of 
unequal length. 

The siphon must be first filled with the liquid to 
be drawn off, and the shorter arm being temporarily 
closed, and then plunged beneath the liquid, a con- 
tinuous flow will take place. 

The action is thus explained : — 

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TI2 



Pneumatics, 



Fio, 




Consider era vertical element of fluid occupying 
the section of the tube at its highest part. 

Then, if a d be a horizontal plane 
continuous with the surface of the 
liquid, the pressure on the side 
A c of the element c r is 

^— c E, 

where ^is the height of a column 
of the same liquid, corresponding 
to the pressure-intensity of the at- 
mosphere; and the pressure on 
the other side is Jf—c f. 

Hence, as c f is greater than c e, there is a resultant 
force acting from right to left, and causing the liquid 
to flow along c d. In this way a continuous flow is 
maintained. It is of course understood that c e is 
not greater than H', for if c e were greater than If^ 
the liquid would commence to flow back along the 
limb c A. Whatever cause may diminish If diminishes 
the maximum height over which the liquid can be 
carried. Should the surface-level of the liquid in the 
vessel fall below b the direction of the resultant force 
would be changed, and the flow would consequently 
cease. 

The resultant force acting on element c r is 
the pressure represented by the column of liquid 

^— CE — (^— Cf) = CF — CE = EF ; 

and, hence, the velocity of efilux is ^/2g ef, 
neglecting friction, and supposing the pressure at b to 
equal the pressure at a. 



d by Google 



A tmospheric Pressure, 1 1 3 

■' ^ ^ Exercises. IX. 

1. If in ascending a mountain the barometer falls from 
76 cm. to 51 cm. find the decrease in pressure on an area of 
one square metre. 

2. A mercury barometer stands at 29*5 in., and the specific 
gravity of mercury is 13*6 : find the specific gravity of oil, if a 
column 36 ft. 6 in. in height can be supported by the at- 
mospheric pressure alone. 

3. When the ordinary mercury barometer stands at 30 in. 
find the whole atmospheric pressure on a surface the area of 
which is 10 square feet. 

4. At what depth below the surface of a lake will the baro- 
nieter indicate a pressure of 50 in. , when the pressure of the 
atmosphere is 30 in. P**^ * 

5. In a siphon barometer of imiform bore the level of the 
mercury in the open end falls through 4 mm. : what change of 
pressure does this indicate ? i 

6. If the sectional areas of the open and closed branches of 
a siphon barometer are as 4 to i, through what distance will 
the mercury move in the closed branch, if the mercury in the 
ordinary barometer rises one inch ? 

7. If the specific gravity of air is 0*0013 when the baro- 
meter stands at 76 cm., find its sp. gr. when the barometer 
stands at 58 cm. 

8. Supposing the average barometric height to be 30 in., 
the sp. gr. of mercury 13-6, and of air 0*0013, ^"^^ the 
height of the atmosphere, supposing the density to be uniform 
throughout. 

9. A barometer is observeii to fall ^ of an inch when car- 
ried up 88 feet of vertical height : how much wovdd it fall if 
taken 1 10 wds up a hill rising i in 3^? 

10. If the specific giravify of niercury is 13*6, what ought to 
be the length of a water barometer inclined to the horizon at 
an angle of 60^, the mercury barometer standing at 30 ins. ? 

1 1. If the height of the column in an ordinary barometer is h^ 
and the tube is inclined through an angle of 30® from the ver- 
tical, what will be the barometric reading? 

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1 14 Pneumatics, 

12. A small bead of glass floats in the mercury of an ordinary 
barometer : does it affect the barometric reading ? 
^ 13. Find the greatest height over which water can be car- 
ried by a siphon at the top of a mountain where the mercury 
barometer stands at 50 cm. 

14. Over what height can a liquid whose sp. gr. is j be 
carried by a siphon, when the height of the mercurial barometer 
is hf and sp. gr. of mercury is /? 

15. A cylindrical body 40 inches high floats in water at the 
ordinary atmospheric pressure with 10 inches of its height im- 
mersed, the sp. gr. of air being '0013. The body, with the 
vessel in which it floats, is then placed under a bell in which the 
atmospheric pressure is ten times as great : what part of the body 
will now be immersed ? 

16. If the atmosphere press with 15 lbs. on every square 
inch, and the weight of a cubic foot of water be 62 J lbs., find 
the total pressure on a rectangular plane surface, placed verti- 
cally with its upper edge a foot deep in water, and its lower 
edge 3 ft. deep, the rectangular surface being 6 ft. long and 2 ft. 
wide. 

17. Find an equation for determining the internal radius of 
a globe of thickness / and specific gravity f, which will just float 
in air of specific gravity j, when filled with gas the density of 
which as compared with air is d. 

18. Taking the pressure of the atmosphere at 15 lbs. per 
square inch, the height of the salt-water barometer at 30 feet, 
calculate the pressure-intensity at 50 fathoms depth in tons per 
square foot. If this pressure acted for a second on a square yard 
of the stem of a vessel weighing 300 tons, what velocity would 
it communicate, there being no resistance to the motion ? 

XV. Relation betiueen the pressure and volume of a 
gas^ the temperature remaining constant. — 
Boyl^s Law, 

We have seen that the characteristic quality of a 
gas is its expansibility and compressibility. We come 

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Boyle's Law, 



IIS 



FiG» SI, 



now to consider the quantitive relation that exists be- 
tween change of volume and change of pressure, the 
temperature remaining constant. This relation can 
be inferred from a series of experiments, of which 
the following are instances : — 

§ 97. Experiments. — For pressures greater than 
the atmospheric pressi/re, — i. A simple form of ap- 
paratus for these experiments consists of a uniform 
bent tube, having one branch open at the top and 
considerably longer than the other, which 
is closed. The shorter branch is some- 
times fitted with a screw-cap so that the 
pressure of the enclosed gas may be more 
easily regulated. The tube itself is gra- 
duated, or a scale is fixed to each branch. 

If the cap, which should fit au:-tight, 
be first unscrewed, and some mercury 
poured into the tube, the mercury will 
rise to the same level in both branches. 
If now the cap be screwed on again some 
air will be enclosed under a pressure equal 
to that of the external atmosphere. Call 
this pressure 76 cm., and suppose the mercury to 
stand at o in each branch. Pour more mercury into 
the tube, and the level of the mercury will be lower 
in the shorter than in the longer branch. The volume 
of the air confined in the shorter branch is now 
diminished, and its pressure is increased by that of 
the column c d. 

Suppose the air in the tube to have originally 
occupied 20 divisions of the tube, or 20 cm., and that 
mercury has been poured into the tube to a height of 

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o 



ti6 



Pneumatics, 



23 cm. above its former level in the longer branch, 
then the mercury in the shorter branch will stand at 
4 cm. above o, and, consequently, the enclosed air 
will occupy only 16 cm. The increase of pressure, due 
to the difference of level, is therefore 23 — 4 = 19 cm. 
Now, if we examine these numbers we shall find 

20 : id:: 19+76 : 76 
i^e. ::95:76, 

or, 20x76= 16x95, 

which shows that the variation in the volume is in* 
versely as the variation in the pressure ; or, what is 
the same thing, that the product of the volume into 
the pressure is constant. 



V 


H 


h 


H-h 


H-h-^n-g-P 


PV 


150 


34-1 


7*3 


26-8 


1027 


1540 


120 


624 


lO'O 


52-4 


1283 


1539 


10 -o 


90-I 


II-8 


78-3 


154-2 


1542 


V 


io6-o 


127 


93-3 


169*2 


1539 


8-3 


I2I-3 


I3'3 


loS-o 


1839 


1526 


7*4 


I47-0 


14-1 


132*9 


208-8 


1545 


67 


i67'5 


148 


1527 


228-6 


1547 


6-4 


1802 


150 


165-2 


241-1 


1537 




Mean value . . 


1539-3 


F«= volume of air In cu 


b. cents. ; H^ height in c 


ents. of 


merc\iry in open lira 


b ; h — height in closed lim 


b; 759 


a» height of baromet 


er. 





On comparing the results of a number of actual 
experiments, the value of the product of the volume 
into the pressure will be found to vary slightly, the 
extent of the variation depending on the degree of 

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Boyl^s Law, 



\\^ 



accuracy with which the experiments are performed. 
In the above table are found the results of a few 
experiments in which the measurements were roughly 
made, the fractional parts of a centimetre being esti- 
mated by eye. They serve, however, to show how 
nearly constant is the value of P V. 

2. For pressures less than the atmospheric pressure. 
To determine the relation between the change in 
the volume and pressure of a gas, when the gas 
expands, we take a barometer tube fitted with a 
screw-cap, and having opened the tube, partly im- 
merse it in a vessel containing mercury. We then 
close the tube, by means of the screw, and thereby 
enclose a certain quantity of air at the p^^ 
ordinary atmospheric pressure. 'J*he tube is 
now raised vertically upwards, and the en- 
closed air expands, its pressure being di- 
minished by that due to the difference of 
surface-level of the mercury in the tube and 
in the vessel. Thus, if ^ ^ = Fbe the volume 
occupied by the air when the mercury 
stands at the same level in the tube and in 
the vessel — that is, under the ordinary atmos- 
pheric pressure H^ and if a //= V be the 
volume occupied at the pressure H-cd^=^P\ 
then, on reference to the scale, it will be 
found that 

ah : ad:: H—cd: H, 
or, VH^ rP'. 

§ 98. Statement of Boyle's* law.— From experi- 

* Robert Boyle born at Lismore in Ireland, 1626 ; died 1691. 

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nen- ' 



Il8 Pneumatics, 

ments such as these performed with different gases, 
and at different temperatures, a law has been estab- 
lished, which we shall see is not accurately true for any 
gas, but approximates very nearly to the truth for 
those gases which are not easily reducible to the 
liquid state. The law is known as Boyle's law or Mar- 
riotte's law, as these two philosophers are said to 
have arrived independently and by similar experiments 
at the same result. It may be enunciated thus : — 

The volume of a gds varies inversely with its pres- 
sure, when the temperature remains constant. 

Thus, if the volume V change to V\ while the 
pressure changes from/ to/', we have : — 

V : v'::p' \p) or Vp=-vy, 

As the mass of the gas remains the same in the 
experiments already described, it follows that the 
density of the gas must increase as the volume 
diminishes, and vice versL Hence Boyle's law may 
be stated thus : — 

The pressure of a gas is proportional to its density, 
tJu temperature remaining constant. 

For a perfect gas, therefore, we have the following 
relations : — 

V' p d ' 

Another statement of this law, due to Professor 
Rankine, places the law in a very clear light : — 

* If we take a closed and exhausted vessel and 
introduce into it one grain of air, this air will, as we 
know, exert a certain pressure on every square inch of 
the surface of the vessel. If we now introduce a 

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Dalton's L aw, 1 1 9 

second grain of air then this second grain will exert 
exactly the same pressure on the sides of the vessel 
that it would have exerted if the first grain had not 
been there before it. Hence we may state, as the 
property of a perfect gas, that any portion of it exerts 
the same pressure against the sides of a vessel as if 
the other portions had not been there.' * 

§ 99. Dalton's Law.^ — This is an extension of 
Boyle's law for a mixture of different gases. If several 
different gases, which do not act chemically on one 
another, are placed in a vessel, the pressure on the 
sides of the vessel is the sum of the pressures due to 
the different gases. Thus, suppose each of the gases, 
if in the vessel by itself, to exert pressures the intensi- 
ties of which are/i, ^2, /a, &c., respectively, the inten- 
sity of the whole pressure exerted by the mixture is 
P\ -^ Pz 4-/3 + &c. Rankine's statement of Boyle's 
law shows this to be the case for different parts of 
the same gas ; hence Boyle's law may be considered 
as a particular case of Dalton's,^ which may be thus 
stated : — When a mixture of several gases ^ at the same 
temperature^ is contained in a vessel, each produces the 
same pressure c^ if the others were not present, 

§ 100. Examples.— (I.) In a bent tube, open at one end 
and closed at the other, the mercury stands at the same level in 
both branches, and the contained air occupies 30 cm. at the 
normal atmospheric pressure — viz. 76 cm. If the section of 
the tube is 10 sq. cm., what volume of mercury must be poured 
into the tube to compress the air to 20 cm. ? 

If / be the pressure of the air when occup3nng 20 cm., wc 



C. Maxwell, Theory of Heat, pp. 27-8. « 1801. 

■ Bom in Cumberland 1766 ; died 1844. 

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I20 Pneumatics, 

have by Boyle's law 20 x /« 30 x 76, .'.;?= 114, of which 76 
is due to atmospheric pressure. Hence the difference of level in 
the two branches equals 38 cm. ; and as the mercury has risen 
10 cm. in the closed branch the quantity of mercury introduced 
equals (20 + 38) ID = 580 cubic cm. 

(2.) The mercury in an ordinary barometer stands at 30 in., 
and the sectional area of the tube is one square inch. A cubic 
inch of air is admitted through the mercury into the vacuum 
above, and depresses the column through 4 inches : find the 
size of the vacuum. • ^ 

When the cubic inch of air is admitted into the vacuum it 
first fills the vacuum, and then with its diminished elasticity, 
consequent on its expansion, it depresses the mercury until the 
pressure of the column of mercury, together with the elasticity 
of the air, equals the atmospheric pressure. 

Suppose the vacuum to measure x inches ; then the air 
which under a pressure of 30 inches occupied i inch is found to 
occupy ;c + 4 inches under a pressure of 30— (30— 4) = 4 in. 

Hence, by Boyle's law, 4 x (^ + 4) = 30, or ;c=3'5 in. 

§ loi. Graphic representation of Boyle's Law. — 

Take o x, o^, two straight lines at right angles to each 
*'^G. 53. other, and along ox 

mark off o a, to repre- 
sent the number of 
units {V) of volume 
which a certain quan- 
_n tity of gas occupies at a 

-^ pressure/. Then, if we 

mark off o b, o c. . . to 
represent 2 ^ 3 F .... we know by Boyle's law that 
the pressures corresponding to these volumes will be 

^, ^ .... Hence, if a p be drawn vertically from 
23* 

A to represent /, then' the lines b q = — , c r = 

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\ 



K 



Graphic Representation of Boyle's Law. 1 2 1 

— , &c. will represent respectively ^^ ?-, &c., and the 

3 . ^. ^ 

curve drawn through these points will serve as a gra- 
phic representation of Boyle's law. 

Now, it is evident that no amount of compression 
can reduce the volume of a gas to zero, and no 
amount of expansion can wholly destroy the pres- 
sure which a perfect gas exerts ; it follows, therefore, 
that the curve will approach on either side nearer and 
nearer to the lines o x and oy, but will never meet 
them. 

Since, also, Vp = V'p' = V"p", we see that the 
product of the volume and the pressure remains con- 
stant, whilst the separate factors vary, and therefore 
the curve p q r is such that if p be any point in it, 
the area of the rectangle o a p a is constant for all 
positions of p. 

Now, if X represent any volume measured along 
ox, and J' the corresponding pressure measured ver- 
tically upwards, we have xy ^=z o. constant, and the 
curve corresponding to this equation is known as the 
rectangular hyperbola. 

§ 102. Graphic representation of the work done 
in changing the volume of a gas from Fto V'^ the 
temperature remaining constant. 

Let a volume V of gas be en- fig. 54. 

closed in the cylindrical vessel d b a c A yy c 
by the piston m n, and let the pis- 



ton be moved through m m' so that ^ ^ w i> 
the volume becomes V, Then if p be the pressure 
of the gas, when the volume is V^ the work done 
in changing the volume to V would be / x m m', 

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122 Pneumatics. 

supposing / to remain the same throughout. But 
this is not the case, for the pressure increases as the 
volume diminishes. If, therefore,/' is the pressure 
of the gas when its volume becomes F', the work 
done is equal to some quantity the value of which 
Ues between / x m m' and />' x m m . 

Now, suppose o a (fig. 55) represent the volume 
V^ and o a' the volume F', a p, a' p' the correspond- 
F»G. 55. ing pressures ; then, if we draw 

the lines p q, p' q', parallel to 
o A, the work done is repre- 
sented by the area of a figure 
which is greater than the rect- 
angle p Q a' A, and less than 
the rectangle p' a' a q'. Now, 
by reasoning similar to that employed for finding the 
space described when a body moves with an increasing 
velocity, it can be shown that the area of the figure 
I p' a' A represents the work done. 

§ 103. Limits of Boyle's Law. — By experiments 
conducted by Regnault ^ and by Despretz* it has been 
observed that Boyle's law is not perfectly true for any 
actual gas. For air and all gases that do not readily 
liquefy under pressure the law is found to be a very 
close approximation to the results of experiments ; 
but for easily liquefiable gases, such as carbonic di- 
oxide and ammonia, the volume is found to decrease 
more rapidly than the pressure increases ; and the 
divergence of the law from the true results is greater 
as the gases approach their point of liquefaction. 

» 1827, 1847. * 1827. 

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Limits of Boyle's Law. 



123 



Thus carbonic-dioxide under a pressure of twenty 
atmospheres is found to occupy only four-fifths of the 
volume given by the law. 

Regnault devised very careful experiments for 
showing the exact relation between the pressure and 
volume of different gases and the extent of the diver- 
gence of this relation from Boyle's law. He arrived 
at the following result : that whereas, according 

to Boyle's law, Vj> = P/, or -/, -1 = 0, the ex- 



Vp 



Vp' 



pression -^J--, — i is a quantity having a small posi- 
tive value for all gases except hydrogen, and increases 
gradually with the pressure. 

If the distances o a, o b, o c, measured along the 
horizontal line o x represent pressures of one, two, 



Fig. 56. 




and three atmospheres, and if the vertical lines a a, 
B b, c c represent the values of ^^ — i for these 

pressures, the curve formed by joining o « ^ ^ . . . is a 
graphic representation of the divergence of the results 
of direct experiment from Boyle's law. 

Seeing that f^ — i is a positive quantity, Vp 

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124 Pneumatics. 

must be greater than V p', and therefore the value of 
V must be less than it would be if Vp = V' p\ in 
accordance with Boyle's law. Hence it appears that 
gases are more compressible than is in accordance 
with Boyle's law. 

In this respect, however, hydrogen differs from 
other gases, and this exception shows that the law is 
more complicated than it would seem to be even with 
the extension above given. It has, however, been 
proved that the true relation of the volume of a gas to 
its pressure depends to some extent on the tempera- 
ture. The curve exhibited in fig. 56 is not the same for 
all temperatures. For nearly all gases the value of 

V V 

the expression -j^ril, — i contmually decreases as the 

V p 

temperature rises, and we are 'thus led to expect 
that if the temperature were sufficiently high this 
quantity would pass through zero and become nega- 
tive ; or the curve, after coinciding with the line o Xy 
would reappear on the other side. It should seem, 
tlierefore, that at a certain temperature varying with 
each gas Boyle's law is strictly accurate, and that 
for higher temperatures the law of the divergence 
is changed, so that the density increases less rapidly 
than the pressure. Now, as hydrogen which has 
already been reduced to the solid state is commonly 
supposed to be the vapour of a metal, it is relatively at 
a very high degree of rarefaction, and this fact may be 

the reason why — ^ - i has a negative value for this 

gas. 

§ 104. Belative Densities of the Air at di£Eerent 

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Variation in Density of Atmosphere, 125 

Heights. — ^We are now in a position to determine the 
law according to which the density of the air changes 
as we ascend from the level of the sea. The decrease 
in the density of air is owing to its compressibility; but 
even if the air were as incompressible as water, and the 
atmosphere were homogeneous throughout, the baro- 
metric column would be found to fall, in rising from 
places of lower to places of higher elevation, in conse- 
quence of the diminution in the height of the column 
of air. In order to obtain an approximate relation 
between the densities of the air at two different heights 
we must neglect the accidental differences of pressure 
caused by differences of temperature and moisture, 
and by the altered value of the force of gravitation. 

Take a vertical column of the atmosphere and sup- 
pose it divided by horizontal planes Fig. 57* 
into a number of strata so thin that 
the density for each layer may be con- 
sidered uniform and equal to that at its 
lower surface. Let the height of the 
column be z^ and let n be the number of _ 



strata, so that the thickness of each layer is - 

n 

let the section of the column be the unit of area. 

Let d and p be the density and pressure of the 
atmosphere at the surface of the earth, and let //j, d<i^ d^ 
• • 'iP\ip2iPz' • • • be the densities and pressures 
at the successive levels. 

Then, since the difference between the pressures 
at the upper and lower surface of a layer must be 
equal to the weight of that layer of air, 

p-px = weight of lowest layer = dg^ (§ 17): 



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126 Pneumatics 

And since by Boyle's law the pre 

density, we have -^ = -^^ -^ = 
a dx d^ 

k {d -- dx)^a 

whence ^/i = (i — S. 

\ k 

similarly ^, = fi^^t 

\ kfi 

/>., the ratio of the densities for t^ 
is constant, since g, k^ z and n are 
fore the quantities d^ d^^ d^, d^ . . 
cal progression, the heights of tl 
arithmetical progression. 

If, therefore, dn be the density 

d \ kn) 

% 105. To find the difference 
stations by means of a baromet 
temperature and force of gravitj 

Let H and h represent the ba 
at the two stations, the vertical dist 
being z^ 

ThenIf:A::d: a 

If V k 



y Google 



Difference of Height and Pressure. 1 27 

where d\% the density of the air at the lower, and d^ the 
density at the upper station. Now, \in increase without 
limit, it is proved in works on Algebra, that the value 

of the right-hand side of this equation becomes e 



Hence ^ = ^ > and \og.S-i ] == f * ^ 
Or,0=^(log.e^-log.e>^) 



Exercises. X. 

1. A gas occupies 100 litres when the barometer stands at 
76 cm, : find the increase in the volume of the gas, if the 
pressure becomes 73 cm. 

2. A tube 2 feet long is filled with water and inverted in a 
vessel of water, with its open end below the surface. Air at 
a pressure of 30 ins. is then admitted into the tube till the level 
of water in the tube is the same as that outside, and the air 
occupies 12 inches. The tube is now raised till the air occupies 
15 inches : find the pressure of the contained air. 

3. Into the vacuum above a common barometer, which stands 
at 30 inches, 2 cubic inches of air are admitted, which depresses 
the mercury 6 inches : if the section of the tube is one square 
inch, find the size of the vacuum. 

4. A vessel of 3 cubic feet capacity containing air at two 
atmospheres' pressure is put into communication with a vessel of 
18 cubic feet capacity containing air at \ of the atmospheric 
pressure : what is now the pressure of the air in the two 
vessels ? 

5. A horizontal cylinder containing air is fitted with a piston 
which is 10 inches from the closed end when the pressure is 
15 lbs. on the square inch. If the area of the piston is 8 square 



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1 28 Pneumatics, 

inches, find the force that must be exerted to hold the piston at 
a distance of 12 inches from the closed end. 

6. If the water barometer stand at 33 feet, to what depth 
must a small cylindrical vessel be sunk to reduce the volume of 
the contained air to one-third of its original volimie, the height 
of the vessel itself being neglected ? 

7. A cylinder contains air at the ordinary atmospheric pres- 
sure, and is fitted with a piston (area 10 square inches), which is 
I ft. from the closed end. If the cylinder be set vertically with 
its open end upwards, how far will the piston descend, the 
atmospheric pressure being 15 lbs. on the square inch, and the 
piston weighing 10 lbs. ? 

8. Suppose the cylinder is held with the open end down- 
wards, how far will the piston fall ? 

9. Find what weight must be hung to the piston in the last 
question to draw it down 2 inches from its original position. 

10. The air contained in a cubical vessel the edge of which is 
one foot, is compressed info a cubical vessel, the edge of which 
is one inch : compare the pressures on the side of each vessel. 

11. Forty c.c. of air are enclosed in a tube over mercury, 
the height of the mercury in the tube above the level in the 
vessel outside being 50 cm. (r ^«5o, fig. 52). The tube is de- 
pressed until cd='}fi cm. What is now the volume of the air, 
the height of the barometer being 76 cm. ? 

12. A bent tube (fig. 51) has a uniform section of I square 
inch and is graduated in inches ; 6 cubic inches of air are en- 
closed in the shorter branch, when the mercury is at the same 
level in both branches. What volume of mercury must be poured 
into the longer branch in order to compress the air into 2 inches ? 
The barometer stands at 30 ins. 

13. The air enclosed in the shorter branch of a similar 
bent tube occupies 11*3 cc, and the difference of level in the 
two branches is 60*2 cm. If the barometer stands at 75*9 cm., 
find what volume the air would occupy under the atmospheric 
pressure only. 



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Experiments oft Diffusion of Gases, 129 



XVI. Diffusion of Gases. 

§ 106. DiflFtision of Oases. Experiments. — We 

have seen that if two layers of different liquids are in 
contact with each other, or are separated by a porous 
diaphragm, a mixture of the liquids takes place, the 
one diffusing into the other. Now, the same pheno- 
menon is observable with gases ; but the laws of 
gaseous diffusion are less complex than those of liquid 
diffusion, in consequence, probably, of the greater 
structural simplicity of gaseous bodies. 

Fill two jars with two different gases — for example^ 
with chlorine and hydrogen — and let the jars be 
connected by a long tube, that containing the hydrogen 
or lighter gas being placed uppermost In a few 
hours the chlorine will find its way into the upper jar, 
as may be seen by its green colour; and the hydrogen 
will take its place. Each of the jars ^^^ ^ 
will be found to contain the same pro- 
portion of the two gases, and the gases 
will remain permanently mixed. This 
intermixture .takes place between any 
gases or vapours which do not act 
chemically on one another. 

If a vessel containing nitrogen be 
covered with a porous diaphragm of 
some colloid substance, and be placed 



under a bell-jar containing hydrogen, /lIU^ 
diffusion will take place, and after a 
time the membrane will have become convex, showing 
that the hydrogen has been passing inwards more 

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1 30 Pfieumatics, 

rapidly than the nitrogen has passed outwards. If the 
position of the gases be reversed the contrary result 
will be found. If the diaphragm be moist, and one 
of the gases is soluble, its rate of diffusion is very 
much increased. Thus, if a moist thin bladder be dis- 
tended with air and placed in a bell-jar containing 
carbonic dioxide, this gas, owing to its solubility, 
passes much more rapidly into the bladder than the 
air escapes from it, and very frequently breaks it, 
though the rate of diffusion of carbonic dioxide into 
air is really less than that of air into the gas. 

§ 107. Rate of DiflFasion. — If we take a long gra- 
duated tube, open at both ends, and close one end 
by a plug of porous clay, and immerse the open end 
in a vessel containing mercury, or water coloured, for 
the sake of greater distinctness, the level of the liquid 
in the tube and in the vessel will be the same, 
showing that the gas must be entering the tube through 
the porous plug at the same rate as 
It escapes mto the outer atmosphere. 
If now we bring an inverted beaker 
filled with coal-gas over the closed 
end of the tube, we find that there is 
a bubbling of gas through the liquid, 
clearly showing that the coal-gas is 
entering the tube more rapidly than 
the air is escaping from it. As soon 
as we withdraw the beaker the liquid 
rises in the tube, the pressure of the 
7 mixed gas in the tube being less 
than that of the atmosphere out- 
side ; and if we replace the beaker 

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Graham's Law. 1 3 1 

the bubbling of the gas through the liquid recom- 
mences. This experiment enables us to observe the 
difference only between the amount of coal-gas that 
goes in and of air that escapes from the tube. We 
can vary this experiment by filling the tube with other 
gases than air. Suppose the tube to be first filled 
with hydrogen and then inverted in the liquid, and 
held in such a position that the level of the liquid 
within and outside the tube is the same, the beaker 
being altogether removed. After a time the liquid 
will be found to rise in the tube, showing that the 
pressure within the tube has become diminished, and 
that the hydrogen must be passing from the tube into 
the external air more rapidly than the air is entering 
the tube. 

It is easy to see that, if the tube be graduated, 
careful experiments will show the volumes of different 
gases that diffuse into the air in the same time. If 
the beaker be filled with a different gas, and held over 
the tube, the rates of diffusion of different gases into 
each other can be ascertained. 

By experiments such as these, Graham established 
the law that the rates of diffusion of two gases into each 
other are in the inverse ratio of the square-roots of their 
densities. For instance, taking the density of air as 
unity ^ that of hydrogen is 0*0692 ; and the square- 
roots of these numbers being i and 0*2632 respectively, 
the law tells us that the rate of diffusion of hydrogen 
is to the rate of diffusion of air 

as I : I-^o•2632, i.e. \\i \ 37994; 

and actual experiment shows that whilst one measure 
K 2 

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132 



Pneumatics, 



of air passes into the tube containing hydrogen, 3*83 
measures of hydrogen escape into the air. 

This important law can be . further verified by 
taking a vessel consisting of two large receivers filled 
with different gases, and connected by a tube with a 
stop-cock. If now the gases be allowed to diffuse into 
each other for a certain period of time, the contents 
of the receivers can be analysed, and the proportion 
of the two gases in each can be quantitively deter- 
mined. By varying the time, and tabulating the 
results, the accuracy of the law may be verified for 
any two gases. 

The following table shows the results of some of 
Graham's experiments : — 



Gas 


Density 


Square 
root of 
density 


z 


Rate of 
diffusion 


VDensity 


Hydrogen .... 
Marsh-gas .... 
Carbonic Oxide . . 
Nitrogen .... 
Oxygen .... 
Nitrous oxide . . . 


0*06926 

0-559 

0*9678 

0*9713 

1*1056 

1*527 


0*2632 
0*7476 
0*9837 
0*9856 
I 0515 
1*0914 


37994 
1-3375 
1*0165 
1*0147 
0*9510 
0*8092 


3-83 

1*34 

I 0149 

1*0143 

0*9487 

082 



If d and d' be the densities of two gases, and D 
and D' the volume of each which diffuses into the 
other in the same time, then, according to the law, 



D:D'::^A=:i ^^, 



I 

\/ d s/d 
OxD^d^D'^d'. 



§ 108. Kinetic Theory of Oases. — In order to ex- 
plain the diffusion of gases we must suppose that the 

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Kinetic Theory of Gases, 133 

particles of a gas axe independent of one another, that 
they are constantly moving in all directions, and with a 
very great velocity. These particles during their motion 
frequently impinge on one another, and the direc- 
tion of their motion is consequently changed. "When 
they come into contact with the sides of the vessel 
containing the gas their momentum is resisted ; and 
it is to this shower of particles moving with a consi- 
derable velocity that the pressure of a gas is supposed 
to be due. The notion that the particles of a gas 
are in rapid motion, and that it is by their impact 
that gases press on one another, is a very old one. It 
was pointed out, not long after Newton's time, by 
Daniel Bemouilli ^ ; Lesage and Prevost of Geneva 
made several applications of the theory, and it was 
afterwards revived in this country by Herepath. In 
1848 Dr. Joule showed how the pressure of gases 
might be explained by the impact of their molecules, 
and he calculated the exact relation that exists be- 
tween the observed pressure of a gas and the velocity 
of its particles. It is, however, to Professors Clausius 
and Clerk-Maxwell that this theory of the molecular 
structure of gases owes its chief development. 

By a method similar to that indicated in Wor- 
meirs * Thermodynamics/ § 71, Joule showed that if a 
vessel contain hydrogen at the ordinary pressure and at 
0° C, the velocity of the particles must be about 6,055 
feet per second. Now, although the velocity of these 
particles is so considerable, the number of particles 
occupying a given volume, say a cubic inch, is so enor- 

* Born at Groningen 1700 ; died 1782. 

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134 Pneumatics. 

mous that the particles move through a very small 
space, and are unable to travel from side to side of the 
vessel containing them, without encountering a series 
of successive impacts with other particles. It follows 
from this, that if the average velocity of the particles 
remains the same, the pressure exerted at any point 
of the vessel containing the gas depends on the num-' 
ber of particles that impinge, in a given time, on the 
element of area containing that point But this, of 
course, depends on the number of particles contained 
in the vessel, />., on the density of the gas ; for if 
the volume of the gas be doubled, the average velocity 
of the particles remaining the same, the number of 
particles traversing a given area will be halved. Now, 
the density of a gas is the ratio iDf its mass to its 
volume, and hence it follows that the pressure a gas 
exerts varies inversely with its volume, if its mass or 
the number of particles in a given volume remain the 
same; and this result is the same as that previously 
obtained by experiment, and known as Boyle's law. 

The notion that a gas consists of a series of parti- 
cles flying about in all directions is the basis of what 
is called the kinetic theory of gases. This theory, 
besides explaining the phenomena of diffusion and 
Boyle's law, accounts for many other facts connected 
with the action of gases at different temperatures; but 
this subject is beyond the range of the present volume. 



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Diving Bell. 135 



CHAPTER VII. 

PNEUMATIC INSTRUMENTS. 

XVII. Diving Bell, Pressure Gauges, 

The two principles which we have now established 
— viz., that the air is a heavy elastic fluid, and that the 
density of a gas varies with its pressure — serve to ex- 
plain most phenomena exhibited by gases at constant 
temperature, and enable us to understand the action 
of a great variety of pneumatic instruments. 

§ 109. The Diving Bell.— If we take an ordinary 
glass tumbler and immerse it vertically, with its mouth 
downwards, in a tub of water, we shall find that 
the enclosed air will prevent the water from rising in 
the inside except to a very small height, which will 
vary with the depth to which the tumbler is immersed. 
The action of the diving-bell is similar. It consists of 
a hollow vessel, nearly cylindrical in form, and open 
at its lower end. When lowered vertically into 
water the enclosed air is compressed by the weight of 
water above it, and the water rises in the bell to a 
height which increases with the depth of the bell from 
the surface of the water. The bell is let down by a 
chain; and in order that men may be able to work 
in the inside, air is introduced by means of a pump 
through an opening in the top, and the pressure of the 
air prevents the water from rising in the bell. 

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136 Pnemnatics, 

§ no. Problems on the Diving Bell.— (i.) To find 
the h'iight to which water rises in the bell, at a given 
depth, when no additional air is introduced : — 

Let B c = 0, be the 
depth of the top of the 
bell below the surface of 
the water. Let b be the 
height of the bell, and H 
the atmospheric pressure 
at the surface of the water 
measured by a water baro- 
meter. 

Then, if c a = x^ the 
part of the bell occupied by the air at the depths, and 
if we suppose the bell to be of uniform area inside, the 
pressure on the air within the bell is equivalent to the 
weight of a column of water the height of which is 
Zr+ BA = ^+^ + ^. Hence, by Boyle's law :— 

X H 





FIO. 


6<>. 




Ma^^ 




WeM 


3£r:^ 


1^ 


i^i^ 


^^^ 




A 


£=£^ 


rrrr.. . 


1^^ 


'■fzzi 



b If+z+x 



or, x^ + X (If + z) = Hb 



a quadratic equation, the positive solution of which 
gives the height required. 

(2.) To find the volume of air at the ordinary 
atmospheric pressure that must be introduced into 
the bell at a given depth, to prevent any water from 
entering: — 

Let z be the depth of the top of the bell, and b the 
height of the bell, as before. Then, if V be the volume 
of the air in the bell at the normal pressure H, and F' 
the volume which the compressed air in the bell at 
depth z + b would occupy at pressure H, it follows 

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Manometers, 137 

from Boyle's law that 

V H 

i\e. the volume of air introduced, at the ordinary 

atmospheric pressure, is T, . V. 
H 

§ III. Manometers, or Pressure-Chkuges. — Mano- 
meters are instruments for measuring the elastic force 
of a gas contained in a closed vessel. fig. 6x. 

The simplest form of manometer con- 
sists of a long narrow open tube which dips 
into a strong box containing mercury. 
The gas, the pressure of which is to be 
measured, is admitted into the box through 
an opening, a, and if the elasticity of the 
gas is equal to that of the air the level of ^ 
the mercury in the tube will be the same | 
as that in the box. If, however, the L 
elasticity of the gas is greater, the mercury will be 
forced up the tube, and the excess of the pressure of 
the gas over that of the atmosphere can be measured 
by the height of the mercury in the tube. 

This form of manometer cannot be used for very 
great pressures, as the length of the tube renders it 
inconvenient. Thus, for a pressure of two atmo- 
spheres the tube must be 30 inches, and the length of 
the tube must be increased 30 inches for every addi- 
tional pressure of one atmosphere that is to be 
measured. 



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138 Pneumatics, 

§ 112. Compressed Air Manometer. — For mea- 
suring greater pressures this form of instrument is 
better suited. 

^^^' ^ _ It consists of a bent tube, one 

branch of which is closed and 
contains air at the ordinary atmo- 
spheric pressure. This air is shut 
off from the other branch by some 
mercury which occupies the 
lower part of the tube. The open 
branch communicates with the 
vessel containing the gas, the pressure of which is to 
be measured. The closed branch of the tube is 
fiunished with a scale. 

If the mercury stands at the same level in both 
branches of the tube the pressure of the gas will equal 
that of the atmosphere. But if a gas or vapour of 
greater pressure be admitted through g the level of 
the mercury will fall in d and rise in c above the ori- 
ginal level, A B. 

The measure of the pressure of the gas in d is that 
of the compressed air in c, together with that indi- 
cated by the difference of level of the mercury in the 
two branches of the tube. 

In order to graduate the scale we must find the 
distance a e or rise of the mercury conesponding 
to a pressure of n atmospheres in d f. Let a c, the 
space originally occupied by the air, equal «, and let 
AE =^. 

Let H be the height of the mercury in the baro- 
meter at the atmospheric prebsuie, and/ the pressure 
of the air in c e. 

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Pressure-gauges. 139 

Then/xcE = ^x ac, or 4 = :i5 = _^ and 
the pressure of the gas in df=/ + 2Ea = n If. 

a—x 
/. 2x'^^ (nH-\-2d)x + {n—i)Ha=^o-y 
and, by the solution of this quadratic, the value of x^ 
corresponding to any number of atmospheres, can be 
determined and the scale graduated. 

§ 113. The Siphon Gauge.— For measuring small 
pressures this instrument is sometimes employed. It 
consists of a bent tube, open at both ends, one of 
which communicates with the vessel containing the 
gas. The liquid used is water or mercury. 

If the gas is admitted through b, and the liquid 
assumes a difference of level, p d, then the pressure of 
the gas equals the atmospheric pres- fig. 63. 
sure + the weight of the liquid in p d. 
If, however, the liquid rises in the 
other branch of the tube, then the 
pressure of the gas = the pressure of 
the atmosphere, minus that due to the 
difference of level. 

Thus, if the liquid fall through x 
inches in one branch, it will rise through 
x inches in the other ; and the difference of level will 
be 2 X. If, therefore, a be the sectional area of the 
tube and s the specific gravity of the fluid — 

The pressure of the gas = atmospheric pressure 
± 2XS, according as the level sinks in b or a. 



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140 Pneumatics, 



Exercises. XI. 

1. To what depth must the top of a diving-bell 8 ft. high be 
immersed under water that the air may be compressed to half its 
volume, the height of the water barometer {H) being equal to 

34 ft.? 

2. What additional volume of air at the ordinary pressure 
(^=34 ft.) must be admitted into a bell 8 feet high, the internal 
section of which is uniform and equal to 20 sq. ft., and the top 
60 feet below the surface, to completely fill it ? 

3. A cylindrical bell, the height of which is 6 feet, is fur- 
nished with a barometer that stands at 30 in., and is lowered into 
water till the barometer stands at 40 in. : find the depth of the 
top of the bell below the surface of the water ; sp. gr. of 
mercury = 13*6. 

4. A barometer marking 30 in. is carried down in a diving- 
bell which is kept constantly full of air : find the depth of the 
top of the bell from the surface of the water, when the barome- 
ter marks 42 in., the height of the bell being 8 feet. 

5. A gas, the pressure of which is 10 lbs. on a square inch, 
communicates with a siphon gauge (fig. 63), the section of which 
is I square inch : find the difference of level, supposing the in- 
strument contains mercury and the ordinary atmospheric pres- 
sure is 15 lbs. on the square inch. 

6. If in the open manometer the distance of the level of the 
mercury from the top of the box is </, and a gas be admitted that 
depresses the mercury b within the box, find the height of the 
mercury in the tube above the original level, the atmospheric 
pressure being h, . 

7. A cylindrical bell, 4 feet deep, whose interior volume is 
20 cubic feet, is lowered into water until its top is 14 feet below 
the surface of the water, and air is forced into it until it is 
three-quarters full. What volume would the air occupy under 
the atmospheric pressure, the water barometer being at 34 feet ? 

8. Find the depth to which a cylindrical diving-bell 8 feet 



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Pumps, 



141 



high must be sunk in water in order that the water may rise in 
it 3 ft. (^= 34 ft.) 

9. If the weight of a cylindrical bell is 2,000 kils. and the 
specific gravity of the material 8, find the tension in the chain 
when the bell is immersed to such a depth that the pressure of 
the enclosed air equals 3 atmospheres, the interior volume being 
I '5 cubic meters. 



Fig. 64. 

c 
P 



XVIII. Air-Pumps. 

§ 1 14. Essential Parts of a Pnmp.— -A pump is an 
instrument for removing a fluid from a reservoir, or 
vessel containing it, by the forcible withdrawal of the 
atmospheric pressure. 

It consists essentially of (i) a barrel or cylinder^ c, 
through which the fluid escapes ; (2) a disc ox piston ^ 
p, capable of moving up 
and down the cylinder, into 
which it exactly fits, and 
worked by a handle or rody 
r, attached to it; (3) a 
pipe^ E, that communicates 
with the reservoir or vessel 
from which the fluid is to 
be removed ; and (4) valves 
or small apertures, with movable covers, opening one 
way only, v, v', which serve to admit the fluid from 
one part of the instrument to another, and to prevent 
it from returning. A valve is generally found at the 
end of the pipe, where it communicates with the 
cylinder, and very frequently in the piston itself. 

§ 115. The Air-Pnmp. — ^This is an instrument for 
removing air from a closed vessel. 

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^ 



142 



Pneumatics. 



Pio. 65. 




It consists of a receiver^ r, from which the air is to 
be removed ; of a cylinder or barrel, a b, furnished 

with a piston and 
valves; and of a pipe 
which serves to con- 
nect the receiver with 
the cylinder. 

The action of the 
pump may be thus 
explained: — 

Suppose the pis- 
ton at B, when the receiver, r, is full of air. As the 
piston is raised the valve m, which opens upwards, re- 
mains closed in consequence of the pressure of the 
external air, and the air from r rushes through the 
pipe E, opens the valve n, and occupies the space 
between the piston and the bottom of the cylin- 
der. When the piston is first pushed down the air in 
the cylinder is compressed, the valve m remaining 
closed j but as soon as the pressure of the air in the 
cylinder begins to exceed that of the air outside the 
valve M is opened, and the enclosed air escapes as the 
piston descends through it. The piston being again 
raised, the same process is repeated. It should be 
noted that it is only when the piston is raised that air 
is withdrawn from the receiver, and that when it 
descends the au: so withdrawn escapes i to the outer 
atmosphere. 

§ 1 16. To determine the density of the air in the 
receiver after any number of strokes. 

Let Fand v be the volumes of the receiver and 
cylinder respectively. Let d be the original density 



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The A ir-Pump, 1 43 

of the air in r, when the piston is at b. Then Vd 
equals the mass of air in r. When the piston rises to 
A the mass of air in r occupies the space V-^-v ; and 
if di be its diminished density we have 

r^=(r+e/)//„or//, = y^ • ^, 

since the mass remains the same. 

When the piston descends to b a part of the air 
escapes, and the mass of air in r = F//i. If now 
d.2 represent the density of the air when its volume 
increases to V + v, we have 

Vd,={V+v) d,, or d, = {y^y d, 

and similarly if d^ represent the density of the air m 
R after three complete strokes — 



Vd^ = 



,^{V-^v) d,, or d, = {jr^' d, 

and, consequently if d^ be the density of the air in 
the receiver after n complete or double strokes 

It will be seen that no amount of exhaustion can re- 
duce the density of the air to zero. At each stroke 
of the piston a fraction only of the air in the receiver 
is removed ; and as the remaining air occupies the 
whole volume of the receiver, a perfect vacuum can- 
not possibly be obtained. 

§ 117. Examples.— ( I.) If the volume of the receiver is 
64 cubic inches and of the cylinder 8 cubic inches, what quan- 

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144 Pneumatics. 

tity of air would be left in the receiver after two complete 
strokes ? 

At the first double stroke 8 cubic inches of air at the original 
density would be removed and 56 would remain. Of these one- 
eighth, or 7 cubic inches, would be removed at the second stroke, 
and consequently 49 would be left. 

(2.) After four complete strokes the density of the air in the 
receiver is to its original density as 10,000 : 14,641 : compare the 
volumes of the receiver and cylinder. 

d 14,641 \K + z// 

F+v V 14,641 II 

§ 118. Difficulty of Workiiig.— It is to be ob- 
served that the difficulty of working this kind of air- 
pump increases with the number of strokes. For, 
neglecting the frictional resistances^ the force required 
to raise the piston depends on the difference of pres- 
sure at ♦the upper and lower surfaces of the piston. 
Now, this difference increases with the rarefaction of 
the air in the receiver, and consequently the difficulty 
of working the machine increases as the exhaustion 
proceeds. 

In lowering the piston the external pressure as- 
sists the action, so long as the air in c is of less den- 
sity than that outside ; but as soon as that point has 
been reached an expenditure of force is necessary to 
overcome the inside pressure and open the valve m. 
Thus the difficulty of lowering the piston decreases 
somewhat with the number of strokes. 

§ 119. The Double-barrelled Air-Pump. — This 
instrument, known as Hawksbee's air-pump, has two 

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The Double-barrelled A ir-pump. 145 



Fig. (>t. 




cylinders, in each of which is a piston worked by 
a rack and pinion. To the 
centre of the toothed-wheel 
is fixed a handle, by moving 
which to and fro the pistons 
are made alternately to as- 
cend and descend. 

The chief advantage of 
this instrument over the one 
already described . is that a 
volume of air equal to that of 
the cylinder is removed at each single stroke of the 
piston, and that, consequently, the rate of exhausting 
the receiver is doubled. 

In the adjoining figure the piston a is ascending 
and withdrawing air from the receiver, and the piston 
B is descending and discharging into the outer atmo- 
sphere the air previously withdrawn from the receiver. 
The position of the valves should be carefully noted. 

Another advantage of this machine is that it is 
easier to work. The difficulty of raising one piston 
is compensated by the assistance which the pressure 
of the atmosphere affords in forcing down the other 
piston, and thus the difficulty of working the machine 
does not increase, as in the case of the single-barrelled 
air-pump, with the number of strokes. In pressing 
down the piston no force is required till the air beneath 
the piston has been compressed to the density of the 
air outside ; and this occurs nearer and nearer to the 
bottom of the cylinder as the degree of exhaustion 
increases. Consequently, the instrument is worked 
somewhat more easily as exhaustion proceeds. 

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146 



Pnmmatics. 



§ 120. Tate's Air-Pump. — This instrument combines 
the advantage of double action with a single barrel. 
The barrel is generally horizontal, and communicates 



Fig. 67. 




with the receiver by a vertical opening, o. The pis- 
ton, c D, occupies a little less than half the length of 
the barrel, and consists generally of two discs rigidly 
connected together by the piston-rod which unites 
both. 

The principle of the action would be the same if 
the piston were solid and of uniform area through- 
out ; but the trouble <A working it would be greater. 
The barrel is furnished with two valves, a and b, at 
either end of it, opening outwards. 

The action may be thus explained : — 
Suppose the piston to be, first of all, in the posi- 
tion shown in fig. i, all the air in firont of it having 
been expelled through a by the driving of the pis- 
ton home. If the piston be now pulled out, as in 

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Tate's Air-pump, 



H7 



fig. 2, the valve a will close, and the air in front of 
the piston will pass through b, which will remain open. 
When the piston reaches the farther end, b, as shown 
in fig. 3, air will escape from the receiver through 
o into the empty space left behind the piston. When 
the piston is pushed inwards this air will be expelled 
through A ; and on the piston reaching a, as shown in 
fig. I, the empty space behind it will be again occupied 
by the air firom the receiver. In this way, a certain 
volume of air, equal to about half the contents of the 
barrel, will be removed at each stroke of the piston ; 
and, as the difference of pressure at the two ends of 
the piston decreases with every stroke, the working 
of the pump becomes easier as the exhaustion pro- 
ceeds. 

§ 121. Mercury Chmge. — The pressure of the air, 
after any number of strokes, in the receiver of an air- 
pump is indicated by a gauge, which is fig. 68. 
generally attached to the connecting- = 
pipe of the air-pump. 

In its simplest form it consists of a 
straight tube, open at both ends. The 
upper end is connected with the receiver, 
and the lower end dips into a cup of 
mercury. As the air is removed firom 
the receiver the pressure inside the tube 
is less than that on the mercury in the 
cup, and consequently the mercury rises "" 
in the tube. This instrument enables us to watch the 
process of exhaustion from the very first stroke. If 
the barometer at the time marks 30 inches, and the 
mercury has risen 4 inches in the tube, the pressure 
L 2 

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14^ 



Pneumatics. 



Fig. 69. 



of the air in the receiver is 30— 4 = i6 in. The 
necessary length of this form of gauge renders it 
somewhat inconvenient. 

The more commonly employed gauge consists of a 
bent tube (fig. 69), having one end closed and the other 
open. Each branch is about 
ten inches in length, and the 
closed end is filled with mer- 
cury, the weight of which is 
supported by the atmospheric 
pressure. The instrument is 
I gg II enclosed in a glass case, which 

_!lL I I communicates with the re- 

ceiver of the air-pump. The 
first few strokes do not produce 
any change in the gauge, but as 
soon as the tension of the au: in the receiver is less 
than the pressure due to the column of mercury in the 
closed end of the tube the mercury begins to fall, and 
the difference of level of the mercury in the two ends 
measures the pressure of the air in the receiver. The 
open end is SQmetimes bent again upon itself, as in 
fig. 70, and screwed into the connecting pipe. 

§ 122. Experiments. — ^The experiments that can 
be performed with the air-pump are very numerous. 
We have found it necessary to refer to some of them, 
in order to prove that the air has weight (§ 90). The 
following additional experiment should also be per- 
formed : — Take a hollow cylindrical vessel and stretch 
a bladder over one end, and then place it on the 
plate of the air-pump, having carefully greased the 
^dges of the glass vessel, so as to prevent the entrance 

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Magdeburg Hemispheres. 



149 



of the air. As the air from ^ under the bladder is 
gradually removed the bladder will be found to yield 
under the pressure of the external air, and will at 
length break. 

By means of the Magdeburg hemispheres, in- 
vented by Guericke, some idea may be formed of the 
magnitude of the pressure which the atmosphere 
exerts on a comparatively small surface. Their gene- 



FlG. 71. 





ral form is shown in fig. 71. The edges are greased 
and pressed together, and the enclosed air is then 
removed by the air-pump. In order to separate the 
two hemispheres by pulling them asunder, a very 
great effort must be made ; and this will be found to 
be the case in whatever position the hemispheres may 
be held, showing also that the atmospheric pressure 
acts equally in all directions. 



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150 Ptieiimatics, 

§ 123. Sprengers Air-Pump. — By means of the 
following contrivance a small receiver can be more 
effectually exhausted than by any of the pumps 
already described. 

It consists of a tube, f b, open at both ends, and 
fitted to a funnel a, by a piece of indiarubber. The 
funnel contains mercury, the flow of 
Fig. 72- which through the tube can be regu- 

^Sil lated by tightening the indiarubber 

▼^ connection. The tube is consider- 

ably longer than a barometer tube, 
and has a spout in its side attached 
to the receiver, e, which is to be ex- 
hausted. The lower end of the tube, 
B, dips into a vessel of mercury. As 
soon as the mercury begins to flow ex- 
haustion commences. The first drops 
of mercury that run out close the 
A lower end of the tube and prevent 

B M. the air from entering. As each drop 

fiy^^^ of mercury passes the neck, c, the 
jr|WS|^. air in the receiver, e, expands and 
^''^^^^^^ occupies the space cq. In this 
way the tube becomes filled with cylinders of air and 
mercury, which gradually escape firom the spout into 
the vessel of mercury. The mercury is poured again 
into the funnel, a, and the process is repeated till the 
tube is occupied by a continuous column of mercury, 
the height of which is equal to that of the mercury 
barometer. The exhaustion is now complete, and 
the receiver, e, corresponds to the Torricellian vacuum 
gf the Qrdinary barometer. 

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Sprengers A ir-pump, 151 

§ 134* ^^ Condensing S]rrmge.-— This consists of 
a barrel, a b, into which an air-tight piston fits, having 
in it a valve that opens downwards. At the bottom 
of the barrel is a valve that likewise opens 
downwards and communicates with a re- ^^°* ^3- 
ceiver, to which the syringe is tightly 
screwed. When the piston is moved 
down, the valve b is closed and the valve 
A opened by the increased pressure. As 
the piston returns, the pressure of the air 
in the receiver closes the valve a, and 
thus prevents the air from re-entering the 
barrel It is evident that the same 
quantity of air will enter the receiver at 
each stroke of the piston, if the machine works per- 
fectly and the piston is moved through the whole 
length of the barrel. 

Exercises. XII. 

1. If the volume of the receiver of an air-pump be eight 
times that of the barrel, compare the density of the air after the 
third stroke with its original density. 

2. If the receiver of an air-pump holds 90 grains of air at 
the ordinary pressure, and if the barrel can hold 10 grains, what 
will be the weight of the air in the receiver after four complete 
strokes ? 

3. If one-third of the contents of the receiver is removed at 
each complete stroke, and the barometer stand at 75 cm., find 
the height to which the mercury will rise in the simple barometer 
gauge after three strokes. 

4. The mercury rises in the barometer gauge through 2^ in. 
in two complete strokes : compare the size of the barrel with 
that of the receiver {h = 30). 

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152 Pneumatics, 

5. The brianches of a siphon barometer gauge are each 
16 cm. long, and the closed branch is filled with mercury, 
which also occupies I cm . of the open branch : compare the den- 
sity of the air in the receiver with its original density, when 
the mercury begins to fall in the siphon-gauge (^ = 75 cm.). 

6. A receiver attached to an air-pump has the volume of 
100 cubic inches, while the cylinder has the volume of 10 cubic 
inches. What proportion of the original air will be left in the 
receiver after the completion of the fourth double stroke? 

7. The capacity of the barrel of a condensing air-pump is 
10 cubic inches, and of a copper receiver 100 cubic inches. By 
how much will the pressure of the air in the receiver be in- 
creased after 20 strokes of the piston ? 

8. When the height of the barometer is 75 cm.^ the air in 
the receiver of an air-pump is exhausted until the mercury in 
the barometer-gauge (fig. 68) attached to it rises from o to 
36 cm. By how much has the tension of the enclosed air been 
reduced ? 



XIX. Pumps for Liquids, 

% 125. Common or Suction Pump.— This is an 

instrument for drawing water from a well or subter- 
ranean reservoir. 

It consists of a cylinder fitted with a piston and 
valve, and connected by a second valve with a pipe 
which communicates with the reservoir. The cylin- 
der M N is called the pump-barrel ; the tube v e the 
suction-tube. The mode of action is as follows : — 

Suppose the piston at first to be at n, and the 
suction-tube filled with air at the atmospheric pres- 
sure. If the piston be raised the air in n d will ex- 
pand, open the valve v, and follow the piston. At the 
same time, since the pressure on the surface of the water 



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The Cofftmon Pump. 



153 



Fig. 74. 




Within the tube is diminished, ovAng to the expansion 

of the contained air, the external pressure at d will 

cause the water to rise in e to such a 

height that the elasticity of the air below 

A B, together with the weight of the column 

of water above D in the suction-tube^ 

equals the ainiospheric pressure without. 

As the piston descends the air below 
it is compressed, and escapes after a 
time through the valve f, the valve v 
being closed by the increased pressure 
of the air above it. Thus the water 
remains at the same level in the suction- 
tube whilst the piston is descending. 

When the piston is again raised the 
pressure is removed from above v, and 
the air underneath it at once opens the 
valve and occupies the space beneath 
A B, the water rising in the suction-tube as before. 

This action continues till the water has risen to n, 
when the raising of the piston causes the water to 
enter the barrel, provided the height d n is less than 
Ji^y the height of the water barometer. As the piston 
continues to rise the water will follow it so long as 
the height of the piston above the level of the water 
outside is less than H, 

As the piston descends, the pressure of the water 
beneath it opens the valve f, and the piston passes 
through the water. When the piston again ascends 
the water is discharged at the spout, and the barrel is 
refilled through the suction-tube. 

The water having once entered the barrel, the 

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154 Pnmmatics. 

contents of the barrel are discharged at each upward 
stroke of the piston \ but, in order that a volume of 
water equal to that of the barrel may be discharged 
at each upward stroke, it is necessary that the atmos- 
pheric pressure should first raise the water to the 
spout ; — i,e, the height of the spout above the level of 
the water outside must be less than H, 

It is e\4dent that if the piston in descending does 
not reach v, so as ultimately to exclude all air from 
underneath it, the water may never be able to enter 
the pump-barrel, even if n d is less than H, 

It is not essential to the working of the pump 
that the tube should be straight ; nor does it matter 
at what horizontal distance from the pump-barrel the 
suction-tube enters the reservoir. 

§ 126. Force required to raise the piston-rod. 

First Suppose the water has not yet entered the 
barrel. In this case the force necessary to raise the 
piston-rod is equal to the difference of the pressure 
of the air on the top and bottom of the piston a b 
(fig. 74). Let E represent the pressure of the air be- 
low A B measured in water, and let ZTbe the height 
of the water barometer outside. Let z equal the 
height of the column of water in the tube. 

Then H^=^ E •\- z\ and the force required to mise 
the piston is 

A B X {H— -^) = A B v. z 
= the weight of a column of water that has a b for 
base, and the distance of surface-level of water in the 
pump above the level outside for height. 

Secondly, Suppose the barrel abeady fiill of water. 
The force required is, as before, the difference of pres- 
sure on the two sides of the piston- r- i 

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The Common Pump. 155 

Let jp = c A (fig. 75) be tlie height of the water 
above the piston at any instant ; then the pressure on 
the upper surface of the piston is p^^ 

(Zr4-^)xAB 
and the pressure on the lower surface is 

(ZT— <s;)x A B 
where z is, as before, the height of the water 
below the piston above the level of the 
reservoir. Hence, the difference, or force 
required is 

AB X {x->rz\ 

or the weight of the column of water having 
A B for base and the difference of the level of the water 
in the pump and in the reservoir for height. This, 
therefore, is the measure of the tension of the piston- 
rod in all cases. 

When the pump is in full action, discharging at 
each stroke a volume of water equal to that of the 
barrel, the tension of the piston-rod is constant 

§ 127. Ezamples. (i.) The length of the suction-tube of a 
common pump is 12 feet, and the piston when at its lowest point 
is 2 feet from the fixed valve ; if at the first stroke the water 
rises 1 1 feet in the tube, find the extreme length of the stroke, 
supposing the water barometer to stand at 33 feet, and the area 
of the barrel to be three times that of the tube. 

If jf be the length of the stroke, and a the area of the tube, 
the volume originally occupied by the air is 12a + 2 x 3^ = i8<j ; 
the volume occupied by the air after the first stroke is 

3a (Jf + 2) + «=(3*+7) a. 

,-. by Boyle's law, -il- «33zJ[i=?. 
3^+7 33 3 
or% Jf=6|feet. 

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1 56 Pneumatics, 

(2.) The suction-tube of a common'pump is 12 feet long, and 
the piston, starting from the fixed valve, is raised at the first 
stroke through 3 feet. If the area of the barrel is four times that 
of the tube, find the height to which the water will be raised in 
the tube {H = 34 feet). 

Let X be the required height, a the area of suction-tube. 
The volume originally occupied by the air was 12a, After the 
first stroke the air occupies (12— jr) a + 3 x 4/7 = (24— jr) a ; and 
the pressure is reduced from 34 to 34— ^ feet. 

Hence, by Boyle's law. 



12 



. 5£_f , or jc « 8*2 feet nearly 



M-x 34 

§ 128. The Liftmg Pomp. — This is a modification 
of the common pump, in which the water discharged 
from the pump-barrel, which is closed at the top, en- 
ters a pipe furnished with a valve and communicating 
with the spout. 

The action is the same as before ; but the water, 
instead of flowing away through the spout, is lifted 
into the pipe and prevented from returning by a valve. 
In this way water can be stored up at any elevation, 
and can afterwards be made to flow through the spout 
as required. 

§ 129. The Forcing Pump.— In this pump the 
piston has no valve. The action is the same as in 
Fig. 76. the common pump until the water enters 
the barrel. Then, as the piston is raised 
the water occupies the space beneath it, 
and passing through the valve c, rises in 
the tube c d to the same level as in the 
barrel. 

As the piston descends the valve b is 
closed, and all the water contained in the 
barrel {^forced up the pipe and prevented 

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^ The Forcing Pump. 1 17 

from returning by the valve C. In this way water 
can be forced up to any height consistent with the 
strength of the material, or can be made to nse m a 
jet from the upper end of the pipe. No flow, how- 
ever, occurs during the ascent of the piston. 

§ 130. Forcing Pump, with Air-vessel.— To ob- 
tain a continuous stream of water from the top of the 
pipe c D (fig. 76) the water must be 
first admitted into a strong vessel 
containing air. The action being 
the same as before, the piston in 
descending forces the water into the 
vessel E F (fig. 77), and compresses 
the air in the upper part of it. As 
more water is forced into the vessel 
the air is further compressed, and 
the water rises in the tube e d and flows out from d. 
When the piston is drawn up the flow from d would 
cease but for the fact that the air in e f, freed from its 
former pressure, now expands and forces the water 
up the tube, thus causing an unbroken flow from d. 

The elasticity of the air in e f will decrease with 
the escape of water from d, and consequently the 
pressure of the water in the pipe must never be greater 
than the excess of the pressure of the air in e f over 
the ordinary atmospheric pressure. If the height of 
the pipe is inconsiderable, the continuity of flow can 
be easily preserved during the ascent of the piston. 

§ 131. The Fire Engine. — This is a double forcing 
pump connected with an air-chamber. The constancy 
of flow is obtained not only by the air-vessel, but also 
by the alternate action of the two pumps. The pistons 

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158 



Pneumatics. 



are worked by a lever, so that one ascends while the 
other descends. Every time the pistons momentarily 

Fig. 78 




Stop, the elasticity of the air in the air-chamber 
maintains the flow. 

§ 132. Brainah*8 Press. — This is a practical appli- 
cation of Pascal's principle of the equal transmissi- 
bility of fluid pressure, and consists of an apparatus 
very similar to that explained in § 27, with the substi- 
tution of a forcing-pump for the weighted piston. 

In fig. 79, A is a platform which supports the 
substance to be pressed against the strong framework 
B. c and D are two solid cylinders which serve as 
pistons, and e and f are two hollow cylinders con- 
nected by a pipe furnished with a valve v. 

The vessel f communicates with a reservoir of 
water by the pipe h, so that d f h constitutes an ordi- 
nary forcing pump. The piston d is attached to a 
lever, k l m, and the power is applied at m. 

When the instrument is in action the vessels e and 
f are filled with water, and as the piston or plunger d 
descends, it closes the valve z/ and forces the water 
into the vessel e, and raises the cylinder c, with its 
attached platform. 

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BramaKs Press. 



159 



Fig. 79. 



r 


^ 


4- ^ 




1 


\ c 


^kjee! 




IP 




1 


•IE « 



If Q be the pressure produced by the plunger d in 

its descent, this pressure is distributed equally over the 

surface of the cylinder c ; and if A be the area of c, 

A 
and a that of d, then the pressure exerted at c is — C. 

d 

If we suppose P to be the force that must be applied 

at M, to produce the pressure Q at l, then it follows 

from the principle of the lever, that 

Q\ P \\ MK : LK 



or, G = 



M K 

iTk ' 



If, therefore, W represent the force with which the 
cylinder, whose area is A^ is pressed upwards, we have 

W^- . Q = ^ . ^. P; and if ^ and ^ are the 
a a L K 



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i6o Pneufnaiics, 

radii of the two cylinders respectively, and M K = / 
and L K = ^, we have 

F f^ ' c 

This instrument is very extensively employed 
where the application of a considerable pressure 
through a small space is required. The very great 
pressure to which the water is subjected necessitates 
great care in the construction of the collars through 
which the cylinders work, so as to render them per- 
fectly water-tight. 

Exercises. XIIL 

• 

1. The spout of a common pump is i6 feet from the surface 
of the water in the reservoir. The area of the pump-barrel is 
72 square inches : find the tension of the piston-rod when the 
pump is full of water. 

2. The specific gravity of mercury is 13*6, and the height 
of the mercurial barometer is 30 inches. What is the greatest 
height to which water can be raised by means of the common 
pump? 

3. Find the pressure that can be produced by a Bramah's 
press if the areas of the pistons are 8 : i, and if a force of 
10 lbs. is applied at the end of a lever 2 feet long, and at a 
distance of 20 ins. from the point where the piston-rod is at- 
tached to it. 

4. If in the lifting pump the area of the barrel is four times 
as great as that of the pipe, compare the pressures on the two 
valves at the commencement of the third stroke after the water 
has entered the pipe. 

5. If the area of the barrel of the forcing pump be 10 square 
inches, and of the pipe into which the water is forced 2 Square 
inches, find the height to which water can be raised in three com- 
plete strokes, supposing the piston-range to be I foot 

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Pneumatic Instruments. 1 6 1 

6. Find the force that must be applied to the piston-rod, in 
the preceding question, at the beginning of the third downward 
stroke. 

7. If the diameter of the piston of a common pum{> be 
4 inches, and the height of the head of the water in the pump 
18 feet above the well, find what pressure the piston bears, 
taking the weight of i cubic foot of water equal to 62*3 lbs. 

8. The length of the suction-tube is 20 feet, and the entire 
stroke of the piston is 6 feet The piston starts from the bottom 
of the barrel, and at the end of the first stroke the water has 
risen 12 feet in the tube. Compare the area of the barrel with 
that of the tube {H^ 34 feet). 

9. A small strong pump is employed for raising mercury from 
a vessel. The height of the fixed valve is 2 feet above the level 
of the mercury in the vessel, and the spout is 8 in. above the 
valve. When the pump is in full action what part of the con- 
tents of the barrel can be ejected at each stroke of the piston 
(^= 30 in. ) ? 

10. A suction-tube is used for drawing up mercury from 3 
vessel containing it, and the piston is raised 10 inches from its 
lowest point, which is 2 inches from the valve at the bottom ol 
the tube. To what height will the mercury rise in the tube 
{iy=3oin.)? 



Vi 



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MISCELLANEOUS PROBLEMS. 



1. A CUBE of brass, whose edge is 2 inches and specific 
gravity 8, is completely imbedded in a cube of wood, whose 
edge is 3 inches and specific gravity 5. Find the mean specific 
gravity of the whole cube. 

2. When equal volumes of alcohol (specific gravity «= o*8) 
and distilled water are mixed together, the volume of the mix- 
ture (after it has returned to its original temperature) is found 
to fall short of the sum of the volumes of its coastituents by 
4 per cent. Find the specific gravity of the mixture. (Univ. 
Lond. Matric.) 

3. The specific gravity of cast copper is 8*88, and that of 
copper wire is 8*79. "What change of volume does a kilo- 
gramme of cast copper undergo in being drawn out into wire? 
(Univ. Lond., istB. Sc.) 

4. A mixture is made of 7 cubic centimetres of sulphuric 
acid (specific gravity = I '843^ and 3 cubic centimetres of dis- 
tilled water, and its specific gravity when cold is found to be 
1*615. Determine the contraction which has taken place. 
(1st B. Sc. 1874.) 

5. Two liquids are mixed (i) by volume in the proportion 
of I : 4, arid (2) by weight in the proportion of 4 : I. The re- 
sulting specific gravities are 2 and 3 respectively. Find the 
specific gravities of the liquids. 

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Miscellaneous Problems. 163 

6. The specific gravity of a mixture of two different liquids 
being supposed to be an arithmetic mean between those of the 
component liquids ; required the ratio of the volumes of the 
latter contained in the mixture. (Univ. Lond., B.A. 1877.) 

7. Find the pressure on a vertical rectangle, 10 inches long 
and 6 inches broad, immersed in water with its longer sides 
horizontal and with the upper one 2 inches below the surface. 
(One cubic foot of water weighs i,ooo ounces.) (Matric. 1877.) 

8. A vessel in the shape of a pyramid, 5 feet high, and with 
a base 4 feet square, is filled with water. Find the pressure 
upon the base, and account for its being greater than the total 
weight in the vessel. 

9. Find the whole pressure on the lower half of the curved 
surface of a vertical cylinder filled with water, the area of the 
base being I2| square cm. and the height 1*4 decim. 

10. A piston, 6 square inches in area, is inserted into one 
side of a closed cubical vessel, measuring 10 feet each way, 
filled with water : the piston is pressed inwards with a force of 
12 lbs. Find the increase of pressure produced on the entire 
surface of the vessel. (B. A. 1872.) 

11. The pressure at the bottom of a well is four times that 
at the depth of 2 feet ; what is the depth of the well if the 
pressure of the atmosphere is equivalent to 30 feet of water ? 
(Camb. Gen. Exam. 1877.) 

12. A and B are vessels fiiU of water, with circular and 
horizontal bases, 12 inches and 8 inches in diameter respec- 
tively. A is 8 inches, and B is 9 inches high. Compare the 
pressure on the bases. (Cam. Gen. Exam.) 

13. If a piece of wood weighing 120 lbs. floats in water with 
four-fifths of its volume immersed, show what is its whole 
volume (I cubic foot of water weighs 625 lbs.). (Matric. 1872.) 

14. A wine-bottle, which below the neck is perfectly 

M 2 



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164 Hydrostatics and Pneumatics, 

cylindrical and has a flat bottom, Is placed in pure water. It is 
found to float upright, with 4J inches immersed. The bottle is 
now removed from the wj^ter and put into oil, the specific 
gravity of which is 0*915. How much of it will be immersed 
in the latter fluid? (Matric. 1874.) 

15. Two pieces of iron (specific gravity 77) suspended from 
the two scale-pans of a balance, the one in water and the other 
in alcohol of the specific gravity 0*85, are found to weigh 
exactly alike. Find the proportion between their true weights. 
(B. Sc. 1875.) 

16. An inch cube of a substance of specific gravity i '2 is 
immersed in a vessel containing two fluids which do not mix 
The specific gravities of these fluids are i*o and i -5. Find what 
will be the point at which the solid will rest. (Matric. 1876.) 

17. A substance which weighs 14 lbs. in air and 12 lbs in 
water, floats in mercury whose density is 13*6. What propor- 
tion of its volume will be immersed ? 

18. A solid, of which the volume is 1*6 cubic centimetres, 
weighs 3*4 gprams in a fluid of specific gravity 0*85. Find the 
specificgravity and weight of the substance. (B. Sc 1876). 

19. An accurate balance is totally immersed in a vessel of 
water. In one scale-pan some glass (specific gravity 2*5) is 
being weighed, and exactly balances a one-pound weight 
(specific gravity 8*o), which is placed in the other scale-pan. 
Find the real weight of the glass. (Matric. 1875. ) 

20. A right cone, whose weight is W, floats in a liquid, 
vortex downwards, with \ of its axis immersed ; what addi- 
tional weight must be placed on the base of the cone so as just 
to sink it entirely in the liquid. (Woolwich Exam.) 

21. A cube floats in distilled water under the pressure of 
the atmosphere, with four-fifthsof its volume immersed and with 
two of its faces horizontal. When it is placed under a con- 
denser where the pressure is that of ten atmospheres, find the 

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Miscellaneous Problems. 165 

alteration in the depth of immersion (the specific gravity of air 
at the atmospheric pressure being; •cx)i3). (B. Sc.) 

22. At the bottom of a mine a mercurial barometer stands at 
77*4 centimetres ; what would be the height of an oil baro- 
meter at the same place, the specific gravity of mercury being 
13-596, and that of oil 09 ? (Matric. 1876.) 

23. A certain quantity of air at atmospheric pressure has a 
volume of 2 cubic feet, the temperature being 55° Fahr. What 
does the volume of the air become when the pressure is increased 
by one-twentieth, the temperature meanwhile remaining the 
same? (Women*s Exam. 1876.) 

24. A syphon barometer is so constructed that the long 
closed tube has an internal sectional area equal to \ of an inch, 
while the short open tube has an internal sectional area equal to 
J an inch. Find what fall will take place in the long tube of 
this barometer when the true pressure of the air falls one inch. 
(B. Sc. 1875.) 

25. The mercury in a barometer stands at £o inches ; the 
section of the tube measures I square inch, and "the vacuum 
above the mercury 6 cubic inches, as much air is passed up the 
tube as depresses the mercury to 29 inches : what would be 
the space occupied by the air under the atmospheric pressure ? 
(B. Sc. 1872.) 

26. In a tube of uniform bore a quantity of air is enclosed. 
What will be the length of this column of air under a pressure 
of three atmospheres, and what under a pressure of a third of an 
atmosphere, its length under the pressure of a single atmosphere 
being 12 inches? (B. A. 1876.) 

27. A Marriotte's tube (fig. 51) has a uniform section of i 
square inch, and is graduated in inches, 6 cubic inches are 
enclosed in the shorter (closed) limb, when the mercury is at 
the same level in both tubes. What volume of mercury must 
be poured into the longer limb, in order to compress the air 

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1 66 Hydrostatics and Pneumatics. 

into 2 inches ? The barometer stands at 30 inches. (B. Sc. 
1874.) 

28. Two cubic centimetres of air are measured off at atmo- 
spheric pressure. When introduced into the vacuum of a baro- 
meter they depress the mercury which previously stood at 
76 cm., and occupy a volume of 15 cubic centimetres. By how 
much has the mercurial column been depressed? (Matric. 1878.) 

29. A tumbler full of air is placed mouth downwards under 
water, at such a depth that the surface of the water inside it is 
at a depth of 25^^ feet. Compare the weight of a cubic inch of 
air in the tumbler with that of a cubic inch of air outside — the 
barometer standing at 30 inches, and the specific gravity of 
mercury being 13-6. 

30. The contents of the receiver of an air-pump is six times 
that of the barrel. Find the elastic force of the air in the re- 
ceiver at the end of the eighth stroke of the piston, when the 
atmospheric pressure is 15 lbs. to the square inch. (B. A. 
1872.) 



Richard Clay ^ Sons, Limited^ London &• Bungay, 

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